Q: Q3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive

This problem involves the concept of apparent depth due to refraction of light at the interface of two media.\n\n**Part 1: Finding the refractive index of water**\n\n**Given:**\n- Real depth of the needle in water, d_real = 12.5 cm\n- Apparent depth of the needle in water, d_apparent = 9.4 cm\n\nThe relationship between refractive index (n), real depth, and apparent depth is:\n n = Real Depth / Apparent Depth\n\n**Calculation:**\n n_water = 12.5 cm / 9.4 cm\n n_water ≈ 1.33\n\nThe refractive index of water is approximately 1.33.\n\n**Part 2: Finding the distance the microscope has to be moved**\n\nNow, the water is replaced by a new liquid.\n\n**Given:**\n- Real depth of the needle in the new liquid, d'_real = 12.5 cm (same height)\n- Refractive index of the new liquid, n_liquid = 1.63\n\nFirst, we calculate the new apparent depth (d'_apparent) for the liquid.\n d'_apparent = d'_real / n_liquid\n\n**Calculation:**\n d'_apparent = 12.5 cm / 1.63\n d'_apparent ≈ 7.67 cm\n\nThe initial position of the microscope focus was at 9.4 cm from the surface. The new position is at 7.67 cm from the surface. Since the apparent depth has decreased, the microscope must be moved upwards (away from the tank) to refocus.\n\n**Distance to be moved:**\nDistance = Initial apparent depth - New apparent depth\nDistance = 9.4 cm - 7.67 cm\nDistance = 1.73 cm\n\nThe microscope would have to be moved **upwards by 1.73 cm** to focus on the needle again.

Q: Q4. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refr

Light from the bulb can emerge out of the water surface only if the angle of incidence at the water-air interface is less than or equal to the critical angle (θ_c). If the angle of incidence exceeds the critical angle, total internal reflection occurs, and light does not emerge. The light emerging from the water surface forms a circular patch. We need to find the radius of this circle.\n\nGiven:\nDepth of water, h = 80 cm = 0.80 m\nRefractive index of water, n_w = 1.33\nRefractive index of air, n_a = 1\n\n**Step 1: Calculate the critical angle (θ_c).**\nThe critical angle for light going from water to air is given by:\nsin(θ_c) = n_a / n_w\nsin(θ_c) = 1 / 1.33\nsin(θ_c) ≈ 0.751879\nθ_c = arcsin(0.751879) ≈ 48.75°\n\n**Step 2: Determine the radius (r) of the circular area.**\nConsider a right-angled triangle formed by the bulb, the point where light emerges at the critical angle, and the center of the circular patch on the surface. The depth of water (h) is the adjacent side, and the radius (r) of the circular patch is the opposite side to the critical angle (θ_c).\n\ntan(θ_c) = r / h\nr = h * tan(θ_c)\nr = 0.80 m * tan(48.75°)\nr = 0.80 m * 1.1397\nr ≈ 0.91176 m\n\n**Step 3: Calculate the area (A) of the surface.**\nThe area through which light can emerge is a circle with radius r.\nA = πr²\nA = π * (0.91176 m)²\nA = π * 0.8313 m²\nA ≈ 2.611 m²\n\nTherefore, the area of the surface of water through which light from the bulb can emerge out is approximately 2.61 m².

Q: Q6. Double-convex lenses are manufactured from glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal le

This problem requires the application of the Lens Maker's Formula.\n\nGiven:\nRefractive index of glass, n = 1.55\nFocal length of the lens, f = +20 cm (since it's a convex lens)\n\nFor a double-convex lens with both faces having the same radius of curvature, let the magnitude of the radius of curvature be R.\nAccording to the Cartesian sign convention:\nFor the first surface (light incident from left), the center of curvature is to the right, so R₁ = +R.\nFor the second surface, the center of curvature is to the left, so R₂ = -R.\n\nThe Lens Maker's Formula is:\n1/f = (n - 1) * (1/R₁ - 1/R₂)\n\n**Step 1: Substitute the given values and sign conventions into the formula.**\n1/20 = (1.55 - 1) * (1/R - 1/(-R))\n\n**Step 2: Simplify the expression.**\n1/20 = (0.55) * (1/R + 1/R)\n1/20 = (0.55) * (2/R)\n1/20 = 1.10 / R\n\n**Step 3: Solve for R.**\nR = 1.10 * 20\nR = 22 cm\n\nTherefore, the radius of curvature required for each face of the double-convex lens is 22 cm.

Q: Q7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens

This problem deals with the concept of a virtual object. When a beam of light converges at a point P, and a lens is placed in its path, the point P acts as a virtual object for the lens. This means the object distance (u) will be positive.\n\nGiven:\nDistance of point P from the lens = 12 cm.\nSince P is a virtual object for the lens, the object distance u = +12 cm.\n\nWe will use the lens formula: 1/v - 1/u = 1/f\n\n**(a) Convex lens of focal length 20 cm**\nGiven:\nFocal length of convex lens, f = +20 cm (positive for convex lens)\nObject distance, u = +12 cm\n\n**Step 1: Apply the lens formula.**\n1/v - 1/u = 1/f\n1/v - 1/(+12) = 1/(+20)\n\n**Step 2: Solve for 1/v.**\n1/v = 1/20 + 1/12\nTo add the fractions, find a common denominator, which is 60.\n1/v = 3/60 + 5/60\n1/v = 8/60\n1/v = 2/15\n\n**Step 3: Calculate v.**\nv = 15/2\nv = +7.5 cm\n\nThe positive sign for v indicates that the image is real and formed on the right side of the lens. So, the beam converges at a point 7.5 cm to the right of the convex lens.\n\n**(b) Concave lens of focal length 16 cm**\nGiven:\nFocal length of concave lens, f = -16 cm (negative for concave lens)\nObject distance, u = +12 cm\n\n**Step 4: Apply the lens formula.**\n1/v - 1/u = 1/f\n1/v - 1/(+12) = 1/(-16)\n\n**Step 5: Solve for 1/v.**\n1/v = -1/16 + 1/12\nTo add the fractions, find a common denominator, which is 48.\n1/v = -3/48 + 4/48\n1/v = 1/48\n\n**Step 6: Calculate v.**\nv = +48 cm\n\nThe positive sign for v indicates that the image is real and formed on the right side of the lens. So, the beam converges at a point 48 cm to the right of the concave lens.

Q: Q9. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

The magnifying power of the telescope depends on whether it is in normal adjustment or if the final image is formed at the least distance of distinct vision (LDDV).

Q: Q10. An object of size 3.0 cm is placed at 14 cm in front of a concave mirror of focal length 21 cm. At what distance from the mirror should a screen be placed to get a sharp focussed i

To determine the image position, size, and nature, we use the mirror formula and magnification formula.\n\nGiven:\nObject size, h_o = +3.0 cm\nObject distance, u = -14 cm (negative as per sign convention for real object in front of mirror)\nFocal length of concave mirror, f = -21 cm (negative for concave mirror)\n\n**1. Position of the image (v):**\nUsing the mirror formula: 1/f = 1/v + 1/u\n1/v = 1/f - 1/u\n1/v = 1/(-21 cm) - 1/(-14 cm)\n1/v = -1/21 + 1/14\nTo add these fractions, find a common denominator, which is 42.\n1/v = (-2/42) + (3/42)\n1/v = 1/42\nv = +42 cm\n\nSince v is positive, the image is formed behind the mirror. A screen cannot be placed behind the mirror to obtain a real image. Therefore, the image is virtual. This means a screen cannot be placed to get a sharp focused image, as the image is virtual and cannot be projected onto a screen.\n\n**2. Size of the image (h_i):**\nUsing the magnification formula: m = h_i / h_o = -v / u\nh_i = h_o * (-v / u)\nh_i = (3.0 cm) * (-(+42 cm) / (-14 cm))\nh_i = (3.0 cm) * (-42 / -14)\nh_i = (3.0 cm) * (+3)\nh_i = +9.0 cm\n\nSince h_i is positive, the image is erect.\n\n**3. Nature of the image:**\nBased on the calculations:\n* Image distance v = +42 cm (positive) indicates the image is formed behind the mirror, hence it is **virtual**.\n* Image size h_i = +9.0 cm (positive) indicates the image is **erect**.\n* Since |m| = 3 > 1, the image is **magnified**.\n\nTherefore, the image is virtual, erect, and magnified. A screen cannot be placed to obtain this image.

Question 1

Q1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to

Accepted Answer

To find the position, nature, and size of the image, we will use the mirror formula and the magnification formula. The effect of moving the candle closer is analyzed by considering how the image distance changes with the object distance.

**Given:**
- Size of the candle (object height), h = +2.5 cm
- Object distance, u = -27 cm (by sign convention)
- Radius of curvature, R = -36 cm (for a concave mirror)

**Calculations:**
1.  **Focal length (f):**
    f = R / 2 = -36 cm / 2 = -18 cm

2.  **Image distance (v):**
    Using the mirror formula: 1/f = 1/v + 1/u
    1/v = 1/f - 1/u
    1/v = 1/(-18) - 1/(-27) = -1/18 + 1/27
    1/v = (-3 + 2) / 54 = -1/54
    v = -54 cm
    The screen should be placed 54 cm in front of the mirror.

3.  **Nature and Size of the image:**
    Magnification, m = -v/u = h'/h
    m = -(-54) / (-27) = -2
    Since magnification 'm' is negative, the image is **real and inverted**.
    Since |m| > 1, the image is **enlarged**.

Image size, h' = m × h
    h' = -2 × 2.5 cm = -5.0 cm
    The size of the image is 5.0 cm. The negative sign indicates it is inverted.

4.  **Effect of moving the candle closer:**
    If the candle is moved closer to the mirror, the object distance 'u' decreases (its magnitude becomes smaller). According to the mirror formula, for a real image formed by a concave mirror, as the object moves from infinity towards the focus, the image moves from the focus away from the mirror. Therefore, as the candle moves closer to the mirror (from -27 cm towards the focus at -18 cm), the screen has to be moved **away from the mirror** to get a sharp image.

Question 2

Q2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved

Accepted Answer

To find the location and magnification of the image formed by the convex mirror, we will use the mirror formula and the magnification formula. The behavior of the image as the object moves away is a key property of convex mirrors.

**Given:**
- Size of the needle (object height), h = +4.5 cm
- Object distance, u = -12 cm (by sign convention)
- Focal length, f = +15 cm (for a convex mirror)

**Calculations:**
1.  **Image distance (v):**
    Using the mirror formula: 1/f = 1/v + 1/u
    1/v = 1/f - 1/u
    1/v = 1/15 - 1/(-12) = 1/15 + 1/12
    1/v = (4 + 5) / 60 = 9/60 = 3/20
    v = +20/3 = +6.7 cm
    The image is formed 6.7 cm behind the convex mirror.

2.  **Magnification (m):**
    m = -v/u
    m = -(+20/3) / (-12) = 20 / (3 × 12) = 20 / 36 = 5/9
    m ≈ +0.56
    The magnification is +5/9.

3.  **Nature and Size of the image:**
    Since 'v' is positive, the image is **virtual** (formed behind the mirror).
    Since 'm' is positive, the image is **erect**.
    Since |m| < 1, the image is **diminished**.
    Image size, h' = m × h = (5/9) × 4.5 cm = 2.5 cm.

4.  **Effect of moving the needle farther:**
    As the needle is moved farther from the mirror, the object distance 'u' increases (its magnitude increases, i.e., u → -∞). 
    From the mirror formula, 1/v = 1/f - 1/u. As u → -∞, the term 1/u → 0.
    Therefore, 1/v → 1/f, which means v → f.
    So, as the needle is moved farther away from the mirror, the image moves from its current position (6.7 cm) **towards the focus** (at 15 cm behind the mirror) and becomes progressively smaller.

Question 3

Q3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive

Accepted Answer

This problem involves the concept of apparent depth due to refraction of light at the interface of two media.

**Part 1: Finding the refractive index of water**

**Given:**
- Real depth of the needle in water, d_real = 12.5 cm
- Apparent depth of the needle in water, d_apparent = 9.4 cm

The relationship between refractive index (n), real depth, and apparent depth is:
 n = Real Depth / Apparent Depth

**Calculation:**
 n_water = 12.5 cm / 9.4 cm
 n_water ≈ 1.33

The refractive index of water is approximately 1.33.

**Part 2: Finding the distance the microscope has to be moved**

Now, the water is replaced by a new liquid.

**Given:**
- Real depth of the needle in the new liquid, d'_real = 12.5 cm (same height)
- Refractive index of the new liquid, n_liquid = 1.63

First, we calculate the new apparent depth (d'_apparent) for the liquid.
 d'_apparent = d'_real / n_liquid

**Calculation:**
 d'_apparent = 12.5 cm / 1.63
 d'_apparent ≈ 7.67 cm

The initial position of the microscope focus was at 9.4 cm from the surface. The new position is at 7.67 cm from the surface. Since the apparent depth has decreased, the microscope must be moved upwards (away from the tank) to refocus.

**Distance to be moved:**
Distance = Initial apparent depth - New apparent depth
Distance = 9.4 cm - 7.67 cm
Distance = 1.73 cm

The microscope would have to be moved **upwards by 1.73 cm** to focus on the needle again.

Question 4

Q4. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refr

Accepted Answer

Light from the bulb can emerge out of the water surface only if the angle of incidence at the water-air interface is less than or equal to the critical angle (θ_c). If the angle of incidence exceeds the critical angle, total internal reflection occurs, and light does not emerge. The light emerging from the water surface forms a circular patch. We need to find the radius of this circle.

Given:
Depth of water, h = 80 cm = 0.80 m
Refractive index of water, n_w = 1.33
Refractive index of air, n_a = 1

**Step 1: Calculate the critical angle (θ_c).**
The critical angle for light going from water to air is given by:
sin(θ_c) = n_a / n_w
sin(θ_c) = 1 / 1.33
sin(θ_c) ≈ 0.751879
θ_c = arcsin(0.751879) ≈ 48.75°

**Step 2: Determine the radius (r) of the circular area.**
Consider a right-angled triangle formed by the bulb, the point where light emerges at the critical angle, and the center of the circular patch on the surface. The depth of water (h) is the adjacent side, and the radius (r) of the circular patch is the opposite side to the critical angle (θ_c).

tan(θ_c) = r / h
r = h * tan(θ_c)
r = 0.80 m * tan(48.75°)
r = 0.80 m * 1.1397
r ≈ 0.91176 m

**Step 3: Calculate the area (A) of the surface.**
The area through which light can emerge is a circle with radius r.
A = πr²
A = π * (0.91176 m)²
A = π * 0.8313 m²
A ≈ 2.611 m²

Therefore, the area of the surface of water through which light from the bulb can emerge out is approximately 2.61 m².

Question 5

Q5. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the

Accepted Answer

This problem involves the prism formula for minimum deviation, first in air and then in water.

**Part 1: Refractive index of the prism material (in air)**
Given:
Angle of minimum deviation, δ_m = 40°
Refracting angle of the prism, A = 60°

The refractive index (n) of the prism material with respect to air is given by the prism formula:
n = [sin((A + δ_m)/2)] / [sin(A/2)]

**Step 1: Substitute the given values into the prism formula.**
n = [sin((60° + 40°)/2)] / [sin(60°/2)]
n = [sin(100°/2)] / [sin(30°)]
n = sin(50°) / sin(30°)

**Step 2: Calculate the values of sin(50°) and sin(30°).**
sin(50°) ≈ 0.7660
sin(30°) = 0.5

**Step 3: Calculate the refractive index (n).**
n = 0.7660 / 0.5
n = 1.532

So, the refractive index of the material of the prism is approximately 1.532.

**Part 2: New angle of minimum deviation when the prism is placed in water**
Given:
Refractive index of prism material, n_g = 1.532 (from Part 1)
Refractive index of water, n_w = 1.33
Refracting angle of the prism, A = 60°

When the prism is placed in water, the light travels from water into the prism and then out into water. The effective refractive index of the prism material with respect to water (n_gw) is required.
n_gw = n_g / n_w
n_gw = 1.532 / 1.33
n_gw ≈ 1.15188

Let the new angle of minimum deviation be δ'_m.
Using the prism formula again with n_gw:
n_gw = [sin((A + δ'_m)/2)] / [sin(A/2)]

**Step 4: Substitute known values into the formula.**
1.15188 = [sin((60° + δ'_m)/2)] / [sin(60°/2)]
1.15188 = [sin((60° + δ'_m)/2)] / [sin(30°)]
1.15188 = [sin((60° + δ'_m)/2)] / 0.5

**Step 5: Solve for sin((60° + δ'_m)/2).**
sin((60° + δ'_m)/2) = 1.15188 * 0.5
sin((60° + δ'_m)/2) ≈ 0.57594

**Step 6: Find the angle (60° + δ'_m)/2.**
(60° + δ'_m)/2 = arcsin(0.57594)
(60° + δ'_m)/2 ≈ 35.16°

**Step 7: Solve for δ'_m.**
60° + δ'_m = 2 * 35.16°
60° + δ'_m = 70.32°
δ'_m = 70.32° - 60°
δ'_m = 10.32°

Therefore, the new angle of minimum deviation when the prism is placed in water is approximately 10.32°.

Question 6

Q6. Double-convex lenses are manufactured from glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal le

Accepted Answer

This problem requires the application of the Lens Maker's Formula.

Given:
Refractive index of glass, n = 1.55
Focal length of the lens, f = +20 cm (since it's a convex lens)

For a double-convex lens with both faces having the same radius of curvature, let the magnitude of the radius of curvature be R.
According to the Cartesian sign convention:
For the first surface (light incident from left), the center of curvature is to the right, so R₁ = +R.
For the second surface, the center of curvature is to the left, so R₂ = -R.

The Lens Maker's Formula is:
1/f = (n - 1) * (1/R₁ - 1/R₂)

**Step 1: Substitute the given values and sign conventions into the formula.**
1/20 = (1.55 - 1) * (1/R - 1/(-R))

**Step 2: Simplify the expression.**
1/20 = (0.55) * (1/R + 1/R)
1/20 = (0.55) * (2/R)
1/20 = 1.10 / R

**Step 3: Solve for R.**
R = 1.10 * 20
R = 22 cm

Therefore, the radius of curvature required for each face of the double-convex lens is 22 cm.

Question 7

Q7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens

Accepted Answer

This problem deals with the concept of a virtual object. When a beam of light converges at a point P, and a lens is placed in its path, the point P acts as a virtual object for the lens. This means the object distance (u) will be positive.

Given:
Distance of point P from the lens = 12 cm.
Since P is a virtual object for the lens, the object distance u = +12 cm.

We will use the lens formula: 1/v - 1/u = 1/f

**(a) Convex lens of focal length 20 cm**
Given:
Focal length of convex lens, f = +20 cm (positive for convex lens)
Object distance, u = +12 cm

**Step 1: Apply the lens formula.**
1/v - 1/u = 1/f
1/v - 1/(+12) = 1/(+20)

**Step 2: Solve for 1/v.**
1/v = 1/20 + 1/12
To add the fractions, find a common denominator, which is 60.
1/v = 3/60 + 5/60
1/v = 8/60
1/v = 2/15

**Step 3: Calculate v.**
v = 15/2
v = +7.5 cm

The positive sign for v indicates that the image is real and formed on the right side of the lens. So, the beam converges at a point 7.5 cm to the right of the convex lens.

**(b) Concave lens of focal length 16 cm**
Given:
Focal length of concave lens, f = -16 cm (negative for concave lens)
Object distance, u = +12 cm

**Step 4: Apply the lens formula.**
1/v - 1/u = 1/f
1/v - 1/(+12) = 1/(-16)

**Step 5: Solve for 1/v.**
1/v = -1/16 + 1/12
To add the fractions, find a common denominator, which is 48.
1/v = -3/48 + 4/48
1/v = 1/48

**Step 6: Calculate v.**
v = +48 cm

The positive sign for v indicates that the image is real and formed on the right side of the lens. So, the beam converges at a point 48 cm to the right of the concave lens.

Question 8

Q8. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of a telescope depends on the focal lengths of its objective and eyepiece lenses and the final image position.

Given:
Focal length of objective lens, f_o = 140 cm
Focal length of eyepiece, f_e = 5.0 cm
Least distance of distinct vision, D = 25 cm

(a) Magnifying power when the telescope is in normal adjustment (final image at infinity):
In normal adjustment, the final image is formed at infinity. The formula for magnifying power (M) in this case is:
M = f_o / f_e
M = 140 cm / 5.0 cm
M = 28

(b) Magnifying power when the final image is formed at the least distance of distinct vision (25 cm):
When the final image is formed at the least distance of distinct vision, the formula for magnifying power (M) is:
M = (f_o / f_e) * (1 + f_e / D)
M = (140 cm / 5.0 cm) * (1 + 5.0 cm / 25 cm)
M = 28 * (1 + 1/5)
M = 28 * (1 + 0.2)
M = 28 * 1.2
M = 33.6

Therefore, the magnifying power of the telescope is 28 in normal adjustment and 33.6 when the final image is formed at the least distance of distinct vision.

Question 9

Q9. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of the telescope depends on whether it is in normal adjustment or if the final image is formed at the least distance of distinct vision (LDDV).

Question 10

Q9.8. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of the telescope is calculated for two cases:
(a) When the telescope is in normal adjustment (final image at infinity), the magnifying power is 28.
(b) When the final image is formed at the least distance of distinct vision (25 cm), the magnifying power is 33.6.

Question 11

Q10. An object of size 3.0 cm is placed at 14 cm in front of a concave mirror of focal length 21 cm. At what distance from the mirror should a screen be placed to get a sharp focussed i

Accepted Answer

To determine the image position, size, and nature, we use the mirror formula and magnification formula.

Given:
Object size, h_o = +3.0 cm
Object distance, u = -14 cm (negative as per sign convention for real object in front of mirror)
Focal length of concave mirror, f = -21 cm (negative for concave mirror)

**1. Position of the image (v):**
Using the mirror formula: 1/f = 1/v + 1/u
1/v = 1/f - 1/u
1/v = 1/(-21 cm) - 1/(-14 cm)
1/v = -1/21 + 1/14
To add these fractions, find a common denominator, which is 42.
1/v = (-2/42) + (3/42)
1/v = 1/42
v = +42 cm

Since v is positive, the image is formed behind the mirror. A screen cannot be placed behind the mirror to obtain a real image. Therefore, the image is virtual. This means a screen cannot be placed to get a sharp focused image, as the image is virtual and cannot be projected onto a screen.

**2. Size of the image (h_i):**
Using the magnification formula: m = h_i / h_o = -v / u
h_i = h_o * (-v / u)
h_i = (3.0 cm) * (-(+42 cm) / (-14 cm))
h_i = (3.0 cm) * (-42 / -14)
h_i = (3.0 cm) * (+3)
h_i = +9.0 cm

Since h_i is positive, the image is erect.

**3. Nature of the image:**
Based on the calculations:
*   Image distance v = +42 cm (positive) indicates the image is formed behind the mirror, hence it is **virtual**.
*   Image size h_i = +9.0 cm (positive) indicates the image is **erect**.
*   Since |m| = 3 > 1, the image is **magnified**.

Therefore, the image is virtual, erect, and magnified. A screen cannot be placed to obtain this image.

Question 12

Q11. A composite slab consisting of two different materials is placed in front of a concave mirror. The two materials have refractive indices 1.5 and 2.0 and thicknesses 20 cm and 30 cm

Accepted Answer

This problem involves the concept of apparent shift due to a glass slab and image formation by a concave mirror. The light from the object first passes through the slab, reflects from the mirror, and then passes through the slab again to form the final image.

Given:
Focal length of concave mirror, f = -10 cm (negative for concave mirror)
Object distance from mirror, u_actual = -60 cm
Slab 1 (closer to mirror): refractive index n1 = 1.5, thickness t1 = 20 cm
Slab 2 (farther from mirror): refractive index n2 = 2.0, thickness t2 = 30 cm

**Step 1: Calculate the total apparent shift due to the composite slab.**
When an object is viewed through a slab, its apparent position shifts. The shift (s) for a single slab is given by s = t(1 - 1/n). For a composite slab, the total shift is the sum of individual shifts.
Shift due to slab 1, s1 = t1(1 - 1/n1) = 20 cm * (1 - 1/1.5) = 20 cm * (1 - 2/3) = 20 cm * (1/3) = 20/3 cm ≈ 6.67 cm
Shift due to slab 2, s2 = t2(1 - 1/n2) = 30 cm * (1 - 1/2.0) = 30 cm * (1/2) = 15 cm
Total apparent shift, S = s1 + s2 = 20/3 cm + 15 cm = (20 + 45)/3 cm = 65/3 cm ≈ 21.67 cm

**Step 2: Determine the effective object distance for the mirror.**
The object is in air, and light from it passes through the slab before reaching the mirror. The slab makes the object appear closer to the mirror. So, the effective object distance (u_eff) for the mirror is:
u_eff = u_actual - S
u_eff = -60 cm - (-65/3 cm) = -60 cm + 65/3 cm = (-180 + 65)/3 cm = -115/3 cm ≈ -38.33 cm

**Step 3: Calculate the image position formed by the mirror (v_mirror).**
Using the mirror formula: 1/f = 1/v_mirror + 1/u_eff
1/v_mirror = 1/f - 1/u_eff
1/v_mirror = 1/(-10 cm) - 1/(-115/3 cm)
1/v_mirror = -1/10 + 3/115
To add these fractions, find a common denominator, which is 230.
1/v_mirror = (-23/230) + (6/230)
1/v_mirror = -17/230
v_mirror = -230/17 cm ≈ -13.53 cm

This means the mirror forms a real image (I') at approximately 13.53 cm in front of the mirror.

**Step 4: Determine the final image position after light passes back through the slab.**
The image I' formed by the mirror is at 13.53 cm from the mirror. This image acts as a real object for the light passing back through the slab. When light passes from a denser medium (or from a position within the slab) to a rarer medium (air), the apparent position shifts *away* from the slab.
So, the final image position (v_final) will be the mirror image position plus the total apparent shift:
v_final = |v_mirror| + S
v_final = |-230/17 cm| + 65/3 cm
v_final = 230/17 cm + 65/3 cm
To add these fractions, find a common denominator, which is 51.
v_final = (230 * 3)/51 + (65 * 17)/51
v_final = (690 + 1105)/51
v_final = 1795/51 cm ≈ 35.20 cm

The final image is formed at approximately 35.20 cm from the mirror, on the same side as the object.

Question 13

Q12. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0

Accepted Answer

To find the separation between the lenses and the magnifying power, we first need to determine the image formed by the objective lens and then use that as the object for the eyepiece.

Question 14

Q13. A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5.0 cm. If this telescope is used to view a 100 m high tower at a distance of 3 km, w

Accepted Answer

We need to calculate the linear height of the image formed by the objective lens first, and then the angular magnification of the telescope in normal adjustment.

Question 15

Q14. A reflecting type telescope has a large concave mirror for its objective with radius of curvature 120 cm. If a plane mirror is used as secondary mirror which is 20 cm from the obje

Accepted Answer

This problem involves a reflecting telescope, likely a Cassegrain type, where a concave objective mirror forms an image, which is then reflected by a secondary plane mirror towards the eyepiece. The final image is formed at the objective's position.

Question 16

Q15. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of a telescope depends on the focal lengths of its objective and eyepiece, and the final image position. We calculate it for two cases: normal adjustment (image at infinity) and when the image is formed at the least distance of distinct vision.

Question 17

Q16. (a) For the telescope described in Exercise 9.15 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3

Accepted Answer

This question extends the previous problem by asking for the length of the telescope and the image height formed by the objective and the final image height. We will use lens formulas and magnification concepts.

Question 18

Q17. A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and

Accepted Answer

The Cassegrain telescope uses a primary concave mirror and a secondary convex mirror. We need to find the final image position for an object at infinity by applying the mirror formula sequentially for both mirrors.

Question 19

Q18. Light from a point source in air falls on a spherical glass surface (n = 1.5) of radius of curvature 20 cm. The distance of the light source from the glass surface is 100 cm. At wh

Accepted Answer

This problem involves refraction at a single spherical surface. We will use the formula for refraction at a spherical surface to find the image position.

Question 20

Q19. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of a telescope depends on the focal lengths of its objective and eyepiece, and the final image position. We will calculate it for two cases: normal adjustment (image at infinity) and when the final image is formed at the least distance of distinct vision (25 cm).

Question 21

Q20. (a) For the telescope described in Exercise 9.19 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3

Accepted Answer

This question extends the previous problem by asking for the length of the telescope and the size of the images formed. We will use the focal lengths from Q9.19 and apply lens formulas and angular magnification concepts.

Question 22

Q21. A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and

Accepted Answer

The Cassegrain telescope uses a concave primary mirror and a convex secondary mirror. We need to find the final image position by considering successive reflections. First, the primary mirror forms an image of the distant object. This image then acts as a virtual object for the secondary mirror, which forms the final image.

Question 23

Q22. Light from a point source in air falls on a spherical glass surface (n = 1.5) of radius of curvature 20 cm. The distance of the light source from the glass surface is 100 cm. At wh

Accepted Answer

This problem involves refraction at a single spherical surface. We will use the formula for refraction at a spherical surface to find the image position.

Question 24

Q23. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

Given:
Focal length of objective lens, f_o = 140 cm
Focal length of eyepiece, f_e = 5.0 cm
Least distance of distinct vision, D = 25 cm

(a) Magnifying power of the telescope in normal adjustment (final image at infinity):
In normal adjustment, the final image is formed at infinity. The magnifying power (M) is given by:
M = f_o / f_e
M = 140 cm / 5.0 cm
M = 28

(b) Magnifying power of the telescope when the final image is formed at the least distance of distinct vision (25 cm):
When the final image is formed at the least distance of distinct vision, the magnifying power (M) is given by:
M = (f_o / f_e) * (1 + f_e / D)
M = (140 cm / 5.0 cm) * (1 + 5.0 cm / 25 cm)
M = 28 * (1 + 1/5)
M = 28 * (1 + 0.2)
M = 28 * 1.2
M = 33.6

Thus, the magnifying power is 28 in normal adjustment and 33.6 when the final image is formed at the least distance of distinct vision.

Question 25

Q24. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. If this telescope is used to view a 100 m tall tower at a distance of 3 km, w

Accepted Answer

Given:
Focal length of objective lens, f_o = 140 cm = 1.4 m
Focal length of eyepiece, f_e = 5.0 cm = 0.05 m
Height of the tower (object), h_o = 100 m
Distance of the tower (object) from the objective lens, u_o = 3 km = 3000 m
Least distance of distinct vision, D = 25 cm = 0.25 m

**Part 1: Height of the image of the tower formed by the objective lens**
For the objective lens, the object (tower) is at a very large distance compared to its focal length. We can use the lens formula for the objective lens:
1/f_o = 1/v_o - 1/u_o
Since the object is very far (u_o = -3000 m, negative as per sign convention), and f_o = 1.4 m:
1/1.4 = 1/v_o - 1/(-3000)
1/1.4 = 1/v_o + 1/3000
1/v_o = 1/1.4 - 1/3000
1/v_o = (3000 - 1.4) / (1.4 * 3000)
1/v_o ≈ 3000 / (1.4 * 3000)
1/v_o ≈ 1/1.4
v_o ≈ 1.4 m

Alternatively, for a distant object, the image formed by the objective lens is approximately at its focal plane. So, v_o ≈ f_o = 1.4 m.

The magnification produced by the objective lens is m_o = h_i / h_o = v_o / u_o
Height of the image formed by the objective lens, h_i = h_o * (v_o / u_o)
h_i = 100 m * (1.4 m / 3000 m)
h_i = 140 / 3000 m
h_i = 14 / 300 m
h_i = 7 / 150 m
h_i ≈ 0.0467 m = 4.67 cm

**Part 2: Height of the final image of the tower if it is formed at 25 cm**
The final image is formed by the eyepiece. The eyepiece acts as a simple magnifier. The angular magnification of the telescope when the final image is at LDDV is:
M = (f_o / f_e) * (1 + f_e / D)
M = (1.4 m / 0.05 m) * (1 + 0.05 m / 0.25 m)
M = 28 * (1 + 1/5)
M = 28 * 1.2
M = 33.6

The angular magnification (M) is also defined as the ratio of the angle subtended by the final image at the eye (β) to the angle subtended by the object at the objective (α).
M = β / α

For a distant object, the angle subtended by the object at the objective is α = h_o / u_o.
α = 100 m / 3000 m = 1/30 rad

The angle subtended by the final image at the eye is β = h_final / D, where h_final is the height of the final image and D is the distance of the final image from the eye (LDDV).
So, M = (h_final / D) / (h_o / u_o)
h_final = M * D * (h_o / u_o)
h_final = 33.6 * 0.25 m * (100 m / 3000 m)
h_final = 33.6 * 0.25 * (1/30) m
h_final = 8.4 * (1/30) m
h_final = 0.28 m = 28 cm

Alternatively, the final image is virtual and magnified. The angular magnification is M = β/α. The angle subtended by the image formed by the objective at the eyepiece is approximately h_i / f_e. The angle subtended by the final image at the eye is h_final / D. However, this is not a direct linear magnification. The question asks for the 'height of the final image'. This is usually interpreted as the linear size of the virtual image formed at LDDV. The angular magnification M relates the angle subtended by the final image to the angle subtended by the object. The height of the final image is not a straightforward concept for a telescope, as it is a virtual image at a specific distance. If we consider the final image to be formed at D, then the angular magnification of the eyepiece is M_e = (1 + D/f_e). The linear size of the image formed by the objective is h_i. The final image height would be h_i * M_e. However, this is not how telescope magnification is typically calculated for linear size. The most appropriate way is using the angular magnification definition.

Let's re-evaluate the interpretation of 'height of the final image'. The angular magnification M = β/α. If the final image is formed at D, then β = h_final / D. So, h_final = M * α * D.
α = h_o / u_o = 100 m / 3000 m = 1/30 rad.
M = 33.6 (from part b of Q23).
D = 0.25 m.
h_final = 33.6 * (1/30) * 0.25 m
h_final = 1.12 * 0.25 m
h_final = 0.28 m = 28 cm.

Final Answer:
Height of the image formed by the objective lens = 4.67 cm
Height of the final image = 28 cm

Question 26

Q25. A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 m apart. If the radius of curvature of the large mirror is 220 m and th

Accepted Answer

Given:
Distance between the large (primary) mirror and the small (secondary) mirror, d = 20 m
Radius of curvature of the large (concave) mirror, R_1 = 220 m
Radius of curvature of the small (convex) mirror, R_2 = 140 mm = 0.14 m
Object is at infinity.

**Step 1: Image formation by the large (primary) concave mirror**
For an object at infinity, a concave mirror forms an image at its focal point.
Focal length of the large mirror, f_1 = R_1 / 2 = 220 m / 2 = 110 m.
Since the object is at infinity (u_1 = -∞), the image v_1 will be formed at f_1.
v_1 = 110 m (This image is real and formed to the left of the primary mirror, towards the secondary mirror).

**Step 2: Image formation by the small (secondary) convex mirror**
The image formed by the primary mirror acts as a virtual object for the secondary mirror.
The distance of this virtual object from the secondary mirror is:
u_2 = v_1 - d
u_2 = 110 m - 20 m = 90 m
Since this image (which is the object for the secondary mirror) is formed to the left of the secondary mirror, and the secondary mirror is a convex mirror, the object distance for the secondary mirror is u_2 = +90 m (as it's a virtual object for the convex mirror, or if we consider the origin at the secondary mirror, the image from the primary mirror is 90m to its left, so u_2 = -90m if we follow standard sign convention where light travels left to right. However, for virtual object, it's often taken as positive. Let's stick to standard sign convention: light from primary mirror travels towards secondary mirror. The image from primary mirror is formed at 110m from primary. Secondary mirror is at 20m from primary. So the image from primary is 110-20 = 90m to the left of secondary mirror. So u_2 = -90m). Let's re-evaluate the sign convention for virtual object. If the rays converge to a point behind the mirror, it's a virtual object. Here, the primary mirror forms a real image at 110m from itself. The secondary mirror is at 20m from the primary. So, the image from the primary mirror is 110 - 20 = 90m *behind* the secondary mirror (i.e., to the left of the secondary mirror, in the direction of the primary mirror). This means the rays from the primary mirror are converging towards a point 90m behind the secondary mirror. Thus, for the secondary mirror, this is a virtual object located at u_2 = +90 m (positive because it's on the side from which light is incident, but behind the mirror).

Focal length of the small (convex) mirror, f_2 = R_2 / 2 = 0.14 m / 2 = 0.07 m.
For a convex mirror, the focal length is positive, so f_2 = +0.07 m.

Using the mirror formula for the secondary mirror:
1/f_2 = 1/v_2 + 1/u_2
1/0.07 = 1/v_2 + 1/90
1/v_2 = 1/0.07 - 1/90
1/v_2 = (90 - 0.07) / (0.07 * 90)
1/v_2 = 89.93 / 6.3
v_2 = 6.3 / 89.93
v_2 ≈ 0.07005 m
v_2 ≈ 7.0 cm

The positive sign of v_2 indicates that the final image is formed to the right of the secondary mirror (i.e., in the direction of the reflected light, which is usually where the eyepiece is placed in a Cassegrain telescope). This image is real.

Therefore, the final image of an object at infinity is formed approximately 7.0 cm to the right of the small (secondary) convex mirror.

Question 27

Q26. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of the telescope is calculated for two cases:
(a) When the telescope is in normal adjustment (final image at infinity), the magnifying power is 28.
(b) When the final image is formed at the least distance of distinct vision (25 cm), the magnifying power is 33.6.

Question 28

Q27. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. If this telescope is used to view a 100 m high tower at a distance of 3 km, w

Accepted Answer

The height of the image of the tower formed by the objective lens is approximately 4.67 cm. The height of the final image of the tower, when formed at the least distance of distinct vision (25 cm), is 28 cm.

Question 29

Q28. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of the telescope is:
(a) 28, when the telescope is in normal adjustment.
(b) 33.6, when the final image is formed at the least distance of distinct vision.

Question 30

Q29. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

Given:
Focal length of objective lens, f_o = 140 cm
Focal length of eyepiece, f_e = 5.0 cm
Least distance of distinct vision, D = 25 cm

(a) Magnifying power when the telescope is in normal adjustment (final image at infinity):
In normal adjustment, the magnifying power (M) of a telescope is given by:
M = f_o / f_e
M = 140 cm / 5.0 cm
M = 28

(b) Magnifying power when the final image is formed at the least distance of distinct vision (D = 25 cm):
When the final image is formed at the least distance of distinct vision, the magnifying power (M) of a telescope is given by:
M = (f_o / f_e) * (1 + f_e / D)
M = (140 cm / 5.0 cm) * (1 + 5.0 cm / 25 cm)
M = 28 * (1 + 1/5)
M = 28 * (1 + 0.2)
M = 28 * 1.2
M = 33.6

Therefore, the magnifying power is 28 in normal adjustment and 33.6 when the final image is formed at the least distance of distinct vision.

Question 31

Q30. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the separation between the objective and the eyepiece when (a) the te

Accepted Answer

Given:
Focal length of objective lens, f_o = 140 cm
Focal length of eyepiece, f_e = 5.0 cm
Least distance of distinct vision, D = 25 cm

(a) Separation between objective and eyepiece when the telescope is in normal adjustment (final image at infinity):
In normal adjustment, the objective forms the image of a distant object at its focus (F_o). This image acts as an object for the eyepiece, which is placed such that this image coincides with its own focal point (F_e). Thus, the distance between the objective and eyepiece is the sum of their focal lengths.
Separation, L = f_o + f_e
L = 140 cm + 5.0 cm
L = 145 cm

(b) Separation between objective and eyepiece when the final image is formed at the least distance of distinct vision (D = 25 cm):
When the final image is formed at the least distance of distinct vision, the image formed by the objective (I_1) acts as an object for the eyepiece. For the final image to be at D, the image I_1 must be located at a distance u_e from the eyepiece such that the eyepiece forms a virtual image at v_e = -D.
Using the lens formula for the eyepiece: 1/f_e = 1/v_e - 1/u_e
1/u_e = 1/v_e - 1/f_e
1/u_e = 1/(-D) - 1/f_e
1/u_e = -1/D - 1/f_e
1/u_e = -(1/D + 1/f_e)
u_e = - (D * f_e) / (D + f_e)
Since u_e is an object distance, we take its magnitude: |u_e| = (D * f_e) / (D + f_e)
|u_e| = (25 cm * 5.0 cm) / (25 cm + 5.0 cm)
|u_e| = 125 cm^2 / 30 cm
|u_e| = 4.166... cm ≈ 4.17 cm

The image formed by the objective is at its focal length f_o (since the object is distant). So, the distance of the first image from the objective is f_o = 140 cm.
The separation between the objective and eyepiece is the sum of the distance of the first image from the objective and the object distance for the eyepiece:
Separation, L = f_o + |u_e|
L = 140 cm + 4.17 cm
L = 144.17 cm

Therefore, the separation is 145 cm in normal adjustment and approximately 144.17 cm when the final image is formed at the least distance of distinct vision.

Question 32

Q31. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. If this telescope is used to view a 100 m tall tower 3 km away, what is the h

Accepted Answer

Given:
Focal length of objective lens, f_o = 140 cm = 1.4 m
Height of the tower (object), h_o = 100 m
Distance of the tower (object) from the objective, u_o = -3 km = -3000 m (negative sign indicates object is real and on the left)

For the objective lens, since the tower is very far away (3 km is much greater than f_o = 1.4 m), the image formed by the objective lens will be approximately at its focal point.
So, the image distance for the objective, v_o ≈ f_o = 1.4 m.

Alternatively, we can use the lens formula for the objective lens:
1/f_o = 1/v_o - 1/u_o
1/v_o = 1/f_o + 1/u_o
1/v_o = 1/1.4 m + 1/(-3000 m)
1/v_o = 1/1.4 - 1/3000
1/v_o = (3000 - 1.4) / (1.4 * 3000)
1/v_o = 2998.6 / 4200
v_o = 4200 / 2998.6 ≈ 1.4006 m
This confirms that v_o is very close to f_o.

Now, we need to find the height of the image (h_i) formed by the objective lens. We use the magnification formula:
M = h_i / h_o = v_o / u_o
h_i = h_o * (v_o / u_o)
h_i = 100 m * (1.4006 m / -3000 m)
h_i = - (100 * 1.4006) / 3000 m
h_i = - 140.06 / 3000 m
h_i ≈ -0.04668 m

The negative sign indicates that the image is inverted. The height of the image is the magnitude of h_i.
Height of the image = 0.04668 m = 4.668 cm

Therefore, the height of the image of the tower formed by the objective lens is approximately 0.0467 m or 4.67 cm.

Question 33

Q32. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. If this telescope is used to view a 100 m tall tower 3 km away, what is the a

Accepted Answer

Given:
Focal length of objective lens, f_o = 140 cm
Focal length of eyepiece, f_e = 5.0 cm
Height of the tower, h_o = 100 m
Distance of the tower, u_o = 3 km = 3000 m

When the telescope is in normal adjustment, the angular magnification (M) is given by:
M = f_o / f_e
M = 140 cm / 5.0 cm
M = 28

Alternatively, angular magnification can be defined as the ratio of the angle subtended by the final image at the eye (β) to the angle subtended by the object at the unaided eye (α).

Angle subtended by the object (tower) at the unaided eye (α):
For small angles, tan α ≈ α = h_o / u_o
α = 100 m / 3000 m = 1/30 radians

Height of the image formed by the objective lens (h_i) was calculated in Q.31:
h_i ≈ 0.04668 m
This image is formed at the focal plane of the objective, so its distance from the objective is v_o ≈ f_o = 1.4 m.
In normal adjustment, this image also lies at the focal plane of the eyepiece. So, the distance of this image from the eyepiece is f_e = 5.0 cm = 0.05 m.

Angle subtended by the final image at the eye (β):
The final image is at infinity, but the angle it subtends is determined by the angle subtended by the intermediate image (h_i) at the eyepiece.
For small angles, tan β ≈ β = h_i / f_e
β = 0.04668 m / 0.05 m
β ≈ 0.9336 radians

Angular magnification, M = β / α
M = 0.9336 / (1/30)
M = 0.9336 * 30
M = 28.008

Both methods yield approximately the same result. The slight difference is due to rounding in intermediate steps and the approximation v_o ≈ f_o. The direct formula M = f_o / f_e is exact for distant objects in normal adjustment.

Therefore, the angular magnification of the tower when viewed through the telescope in normal adjustment is 28.

Question 34

Q33. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whe

Accepted Answer

The magnifying power of the telescope is calculated for two cases:
(a) When the telescope is in normal adjustment (final image at infinity), the magnifying power is 28.
(b) When the final image is formed at the least distance of distinct vision (25 cm), the magnifying power is 33.6.

Question 35

Q34. (a) For the telescope described in Exercise 9.33 (objective focal length 140 cm and eyepiece focal length 5.0 cm), what is the separation between the objective lens and the eyepiec

Accepted Answer

The separation between the objective lens and the eyepiece depends on the final image position.
(a) In normal adjustment, the separation is 145 cm. When the final image is at the least distance of distinct vision (25 cm), the separation is approximately 144.17 cm.
(b) The height of the image of the tower formed by the objective lens is -4.67 cm (inverted).
(c) The height of the final image of the tower when it is formed at 25 cm is approximately -26.68 cm (inverted).

Question 36

Q35. For the normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of abou

Accepted Answer

The range of accommodation of the eye-lens is the difference between its maximum and minimum converging powers. We can calculate these powers by considering the eye focusing at the far point and the near point.

**Given:**
- Power of cornea, P_c = 40 D
- Least power of eye-lens, P_e(min) = 20 D
- Near point distance, d = 25 cm = 0.25 m
- Far point is at infinity.

**1. To see objects at infinity (Far Point):**
The eye is in its most relaxed state, and the power of the eye-lens is minimum.
Total power of the eye, P_min = Power of cornea (P_c) + Minimum power of eye-lens (P_e(min))
P_min = 40 D + 20 D = 60 D

The focal length of the eye-lens system is f_min = 1 / P_min = 1/60 m.
For an object at infinity (u = ∞), the image is formed on the retina. The distance from the lens to the retina (v) is equal to the focal length.
So, v = 1/60 m ≈ 1.67 cm.
This is the fixed distance between the cornea/lens system and the retina.

**2. To see objects at the near point:**
The eye exerts maximum effort to focus, and the power of the eye-lens is maximum.
Object distance, u = -25 cm = -0.25 m
Image distance, v = 1/60 m (fixed distance to the retina)

Let P_max be the maximum total power of the eye. Using the lens formula in terms of power:
P = 1/v - 1/u
P_max = 1 / (1/60) - 1 / (-0.25)
P_max = 60 + 4 = 64 D

This maximum total power is the sum of the cornea's power and the maximum power of the eye-lens (P_e(max)).
P_max = P_c + P_e(max)
64 D = 40 D + P_e(max)
P_e(max) = 64 D - 40 D = 24 D

**3. Range of Accommodation:**
The range of accommodation of the eye-lens is the difference between its maximum and minimum powers.
Range = P_e(max) - P_e(min)
Range = 24 D - 20 D = 4 D

Thus, the estimated range of accommodation for a normal eye is about 4 dioptres.

Question 37

Q36. Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its power of accommodation? If not, what might cause these def

Accepted Answer

No, short-sightedness (myopia) or long-sightedness (hypermetropia) does not necessarily imply that the eye has lost its power of accommodation. Accommodation is the ability of the eye-lens to change its focal length to focus objects at various distances onto the retina. A person with myopia or hypermetropia usually retains this ability, although the range over which they can see clearly is shifted.

These defects are refractive errors caused by a mismatch between the converging power of the eye's lens system and the length of the eyeball.

**Causes of Myopia (Short-sightedness):**
In a myopic eye, the light rays from a distant object (at infinity) are focused in front of the retina, not on it. This results in blurry vision for distant objects. The causes can be:
1.  **Elongation of the eyeball:** The distance between the eye-lens and the retina becomes too large.
2.  **Excessive converging power:** The cornea or the eye-lens (or both) has a shorter focal length (higher power) than normal, causing light to bend too much.

A myopic person can see near objects clearly, often even closer than the normal near point, because their eye's focusing power is naturally strong. They have a functional power of accommodation.

**Causes of Hypermetropia (Long-sightedness):**
In a hypermetropic eye, the light rays from a distant object are focused behind the retina. This results in blurry vision for near objects, and can cause strain when viewing distant objects. The causes can be:
1.  **Shortening of the eyeball:** The distance between the eye-lens and the retina is smaller than normal.
2.  **Insufficient converging power:** The cornea or the eye-lens has a longer focal length (lower power) than normal, causing light to not bend enough.

A person with hypermetropia has to use their accommodation power even to see distant objects clearly. This continuous strain can lead to headaches. Their ability to accommodate is present, but it is already partially used for distant vision, leaving less reserve for viewing near objects.

**Loss of Accommodation:**
The condition that directly relates to the loss of accommodation power is **Presbyopia**. This is an age-related condition where the ciliary muscles weaken and the eye-lens loses its flexibility, making it difficult to focus on near objects. It is distinct from myopia and hypermetropia, although a person can have both (e.g., a myopic person developing presbyopia in old age).

Question 38

Q37. An object of size 3.0 cm is placed 14 cm in front of a concave mirror of focal length 21 cm. The object is placed perpendicular to the principal axis. Find the position, nature and

Accepted Answer

The image is formed at a distance of 42 cm behind the mirror. The image is virtual, erect, and its size is 9.0 cm.

Question 39

Q38. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive

Accepted Answer

The refractive index of water is 1.33. If water is replaced by a liquid of refractive index 1.63, the microscope would have to be moved downwards by 1.6 cm to focus on the needle again.

Chapter 9: Ray Optics and Optical Instruments