Question 1

1 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 obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Accepted Answer

Given:
Size of candle, h = 2.5 cm
Object distance, u = -27 cm (since object is in front of mirror)
Radius of curvature, R = 36 cm
Focal length, f = R/2 = 36/2 = 18 cm

Using mirror formula:
1/f = 1/v + 1/u
=> 1/v = 1/f - 1/u = 1/18 - 1/(-27) = 1/18 + 1/27 = (3 + 2)/54 = 5/54
=> v = 54/5 = 10.8 cm

So, the screen should be placed at 10.8 cm in front of the mirror to get a sharp image.

Nature of image:
Since v is positive, image is real and formed on the same side as the reflected light.

Magnification, m = -v/u = -10.8 / (-27) = 0.4

Size of image = m × size of object = 0.4 × 2.5 cm = 1.0 cm

The image is real, inverted, and diminished in size.

If the candle is moved closer to the mirror (i.e., u decreases in magnitude), then from the mirror formula, v will increase (image moves away from mirror). Hence, the screen should be moved farther from the mirror to get a sharp image. — Step 1: Calculate focal length from radius of curvature.
Step 2: Use mirror formula to find image distance.
Step 3: Determine magnification and image size.
Step 4: Analyze nature of image from sign conventions.
Step 5: Discuss effect of moving object closer on image position.

Question 2

2 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 farther from the mirror.

Accepted Answer

Given:
Object size, h = 4.5 cm
Object distance, u = -12 cm (object in front of mirror)
Focal length, f = +15 cm (convex mirror focal length is positive)

Using mirror formula:
1/f = 1/v + 1/u
=> 1/v = 1/f - 1/u = 1/15 - 1/(-12) = 1/15 + 1/12 = (4 + 5)/60 = 9/60 = 3/20
=> v = 20/3 = 6.67 cm

Image distance v is positive, so image is virtual and behind the mirror.

Magnification, m = -v/u = -6.67 / (-12) = 0.556

Image size = m × object size = 0.556 × 4.5 cm = 2.5 cm

The image is virtual, erect, and diminished.

As the needle is moved farther from the mirror (increasing |u|), the image distance v approaches the focal length f (15 cm) from the mirror, and the image size decreases further, approaching zero magnification. — Step 1: Use mirror formula with given focal length and object distance.
Step 2: Calculate image distance and magnification.
Step 3: Interpret signs to describe image nature.
Step 4: Discuss effect of increasing object distance on image position and size.

Question 3

3 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 index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Accepted Answer

Given:
Real depth, d = 12.5 cm
Apparent depth, d' = 9.4 cm

Refractive index, $ \mu = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{d}{d'} = \frac{12.5}{9.4} = 1.33 $

If water is replaced by a liquid of refractive index $ \mu' = 1.63 $ and same real depth d = 12.5 cm,

New apparent depth, d'' = $ \frac{d}{\mu'} = \frac{12.5}{1.63} = 7.67 $ cm

Change in apparent depth = d' - d'' = 9.4 - 7.67 = 1.73 cm

Therefore, the microscope must be moved closer by 1.73 cm to focus on the needle again. — Step 1: Use relation between real depth, apparent depth, and refractive index.
Step 2: Calculate refractive index of water.
Step 3: Calculate new apparent depth for liquid with refractive index 1.63.
Step 4: Find difference in apparent depths to determine microscope movement.

Question 4

4 Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].

Accepted Answer

Given:
Angle of incidence in water, i = 45°

From figures 9.27(a) and (b), refractive indices can be estimated or known:
Assuming refractive index of air, n_air = 1
Refractive index of glass, n_glass ≈ 1.5
Refractive index of water, n_water ≈ 1.33

Using Snell's law at water-glass interface:

n_water * sin(i) = n_glass * sin(r)

=> sin(r) = (n_water / n_glass) * sin(45°) = (1.33 / 1.5) * 0.7071 = 0.627

=> r = sin⁻¹(0.627) ≈ 38.9°

Therefore, the angle of refraction in glass is approximately 39°. — Step 1: Identify refractive indices from given figures or standard values.
Step 2: Apply Snell's law at water-glass interface.
Step 3: Calculate sine of refracted angle.
Step 4: Find refracted angle using inverse sine.

Question 5

5 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? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Accepted Answer

Given:
Depth of water, h = 80 cm
Refractive index of water, n = 1.33

At the water-air interface, light undergoes total internal reflection beyond the critical angle.

Critical angle, $ \theta_c = \sin^{-1} \left( \frac{n_{air}}{n_{water}} \right) = \sin^{-1} \left( \frac{1}{1.33} \right) = 48.75^\circ $

Light rays inside water can emerge only within this critical angle.

The radius R of the circular area on the surface through which light emerges is given by:
$$ \tan \theta_c = \frac{R}{h} \implies R = h \tan \theta_c = 80 \times \tan 48.75^\circ $$

Calculating:
$ \tan 48.75^\circ \approx 1.14 $

So,
$$ R = 80 \times 1.14 = 91.2 \text{ cm} $$

Area of surface = $ \pi R^2 = \pi \times (91.2)^2 = \pi \times 8317.44 = 26128 \text{ cm}^2 $

Therefore, the area is approximately 26128 cm² or 2.61 m². — Step 1: Calculate critical angle using refractive indices.
Step 2: Use geometry to find radius of circle on surface.
Step 3: Calculate area of circle using radius.
Step 4: Convert units if necessary.

Question 6

6 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 refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Accepted Answer

Given:
Angle of minimum deviation, $ \delta_m = 40^\circ $
Refracting angle of prism, A = 60°

Using the formula for refractive index of prism material:
$$ n = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} $$

Calculate:
$$ n = \frac{\sin \left( \frac{60 + 40}{2} \right)}{\sin 30} = \frac{\sin 50^\circ}{0.5} = \frac{0.7660}{0.5} = 1.532 $$

So, refractive index of glass, n = 1.532

When prism is placed in water (n_water = 1.33), the refractive index relative to water is:
$$ n' = \frac{n_{glass}}{n_{water}} = \frac{1.532}{1.33} = 1.152 $$

Let new angle of minimum deviation be $ \delta_m' $.

Using the same formula:
$$ n' = \frac{\sin \left( \frac{A + \delta_m'}{2} \right)}{\sin \left( \frac{A}{2} \right)} $$

$$ 1.152 = \frac{\sin \left( \frac{60 + \delta_m'}{2} \right)}{0.5} \implies \sin \left( \frac{60 + \delta_m'}{2} \right) = 0.576 $$

$$ \frac{60 + \delta_m'}{2} = \sin^{-1} 0.576 = 35.2^\circ $$

$$ 60 + \delta_m' = 70.4^\circ \implies \delta_m' = 10.4^\circ $$

Therefore, the new angle of minimum deviation is approximately 10.4°. — Step 1: Use prism formula to find refractive index from minimum deviation.
Step 2: Calculate refractive index of glass.
Step 3: Find relative refractive index when prism is in water.
Step 4: Use formula again to find new minimum deviation angle.

Question 7

7 Double-convex lenses are to be manufactured from a 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 length is to be 20 cm?

Accepted Answer

Given:
Refractive index, n = 1.55
Focal length, f = 20 cm
Both faces have same radius of curvature, R

Lens maker's formula:
$$ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$

For double convex lens with equal radii:
$$ R_1 = R, \quad R_2 = -R $$

So,
$$ \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \left(-\frac{1}{R} \right) \right) = 2 (n - 1) \frac{1}{R} $$

Rearranged:
$$ R = 2 (n - 1) f $$

Calculate:
$$ R = 2 \times (1.55 - 1) \times 20 = 2 \times 0.55 \times 20 = 22 \text{ cm} $$

Therefore, the radius of curvature required is 22 cm. — Step 1: Use lens maker's formula for focal length.
Step 2: Substitute equal radii with opposite signs.
Step 3: Rearrange to find radius of curvature.
Step 4: Calculate numerical value.

Question 8

8 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 of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Accepted Answer

Given:
Distance of lens from point P = 12 cm

(a) Convex lens, focal length f = +20 cm

The beam converges at P, so the image formed by the lens will be at a distance v from the lens.

Object distance u = -12 cm (since beam converges at P, object for lens is virtual and on same side as image)

Using lens formula:
$$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-12} = \frac{1}{20} - \frac{1}{12} = \frac{3 - 5}{60} = -\frac{2}{60} = -\frac{1}{30} $$

$$ v = -30 \text{ cm} $$

Negative v means image is formed 30 cm on the same side of the lens as the object.

Therefore, the beam converges 30 cm on the same side of the lens as the object, i.e., 30 cm before the lens.

(b) Concave lens, focal length f = -16 cm

Using lens formula:
$$ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-16} + \frac{1}{-12} = -\frac{1}{16} - \frac{1}{12} = -\frac{3 + 4}{48} = -\frac{7}{48} $$

$$ v = -\frac{48}{7} = -6.86 \text{ cm} $$

Image is formed 6.86 cm on the same side as the object.

Therefore, the beam converges 6.86 cm before the lens. — Step 1: Identify object distance for the lens (virtual object).
Step 2: Apply lens formula with given focal lengths.
Step 3: Calculate image distance for convex and concave lenses.
Step 4: Interpret sign conventions to find actual convergence points.

Question 9

9 An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Accepted Answer

Given:
Object size, h = 3.0 cm
Object distance, u = -14 cm
Focal length, f = -21 cm (concave lens focal length is negative)

Using lens formula:
$$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-21} + \frac{1}{-14} = -\frac{1}{21} - \frac{1}{14} = -\frac{2 + 3}{42} = -\frac{5}{42} $$

$$ v = -\frac{42}{5} = -8.4 \text{ cm} $$

Image distance is negative, so image is virtual and on the same side as the object.

Magnification, m = v/u = -8.4 / -14 = 0.6

Image size = m × object size = 0.6 × 3.0 = 1.8 cm

Image is virtual, erect, and diminished.

If the object is moved further away (increasing |u|), the image distance v approaches the focal length f from the lens side, and the image size decreases further, remaining virtual and erect. — Step 1: Use lens formula with given focal length and object distance.
Step 2: Calculate image distance and magnification.
Step 3: Interpret signs to describe image nature.
Step 4: Discuss effect of increasing object distance on image position and size.

Question 10

10 What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Accepted Answer

Given:
Focal length of convex lens, f1 = +30 cm
Focal length of concave lens, f2 = -20 cm

For lenses in contact, effective focal length f is given by:
$$ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \left(-\frac{1}{20}\right) = \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} = -\frac{1}{60} $$

$$ f = -60 \text{ cm} $$

Negative focal length means the combination behaves like a diverging lens.

Therefore, the system is a diverging lens with focal length -60 cm. — Step 1: Use formula for effective focal length of lenses in contact.
Step 2: Substitute given focal lengths with correct signs.
Step 3: Calculate effective focal length.
Step 4: Interpret sign to determine lens type.

Question 11

11 A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Accepted Answer

Given:
Focal length of objective, f_o = 2.0 cm
Focal length of eyepiece, f_e = 6.25 cm
Distance between lenses, L = 15 cm

(a) Final image at least distance of distinct vision, D = 25 cm

Let object distance from objective be u_o, image distance from objective be v_o.

Using lens formula for objective:
$$ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} $$

The image formed by objective acts as object for eyepiece.

Distance of this image from eyepiece = L - v_o

For final image at distance D from eye (on same side as object for eyepiece), eyepiece formula:
$$ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} $$

Here, final image distance v_e = -D (since image is virtual and on same side as object for eyepiece), and object distance for eyepiece u_e = L - v_o.

So,
$$ \frac{1}{f_e} = \frac{1}{-D} - \frac{1}{L - v_o} \implies \frac{1}{L - v_o} = \frac{1}{-D} - \frac{1}{f_e} $$

Calculate:
$$ \frac{1}{L - v_o} = -\frac{1}{25} - \frac{1}{6.25} = -0.04 - 0.16 = -0.20 $$

$$ L - v_o = -5 \text{ cm} \implies v_o = L + 5 = 15 + 5 = 20 \text{ cm} $$

Using objective lens formula:
$$ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{20} - \frac{1}{2} = 0.05 - 0.5 = -0.45 $$

$$ u_o = -2.22 \text{ cm} $$

Object is placed 2.22 cm in front of objective.

Magnifying power (M) of microscope:
$$ M = m_o \times m_e = \left( \frac{v_o}{u_o} \right) \times \left( 1 + \frac{D}{f_e} \right) $$

Calculate:
$$ m_o = \frac{v_o}{u_o} = \frac{20}{-2.22} = -9.0 $$

$$ m_e = 1 + \frac{25}{6.25} = 1 + 4 = 5 $$

$$ M = -9.0 \times 5 = -45 $$

Negative sign indicates image is inverted.

(b) Final image at infinity:

For final image at infinity, image formed by objective is at focal point of eyepiece:
$$ L - v_o = f_e = 6.25 \text{ cm} \implies v_o = L - f_e = 15 - 6.25 = 8.75 \text{ cm} $$

Using objective lens formula:
$$ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{8.75} - \frac{1}{2} = 0.114 - 0.5 = -0.386 $$

$$ u_o = -2.59 \text{ cm} $$

Magnifying power:
$$ M = m_o \times m_e = \left( \frac{v_o}{u_o} \right) \times \frac{D}{f_e} = \frac{8.75}{-2.59} \times \frac{25}{6.25} = -3.38 \times 4 = -13.52 $$

Therefore, magnifying power is approximately 13.5 (inverted image). — Step 1: Use lens formula for objective and eyepiece.
Step 2: Calculate image distance for objective lens.
Step 3: Use eyepiece formula for final image at given distances.
Step 4: Calculate magnifications of objective and eyepiece.
Step 5: Multiply to get total magnifying power.

Question 12

12 A person with a normal near point (25cm) 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 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Accepted Answer

Given:
Near point, D = 25 cm
Focal length of objective, f_o = 8.0 mm = 0.8 cm
Focal length of eyepiece, f_e = 2.5 cm
Object distance from objective, u_o = -9.0 mm = -0.9 cm

Using lens formula for objective:
$$ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{0.8} + \frac{1}{-0.9} = 1.25 - 1.11 = 0.14 $$

$$ v_o = 7.14 \text{ cm} $$

Separation between lenses, L = v_o + u_e

Object for eyepiece is image formed by objective, so object distance for eyepiece, u_e = -v_o = -7.14 cm

For final image at near point, image distance for eyepiece, v_e = -D = -25 cm

Using lens formula for eyepiece:
$$ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \implies \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{2.5} = -0.04 - 0.4 = -0.44 $$

$$ u_e = -2.27 \text{ cm} $$

But this contradicts previous u_e = -7.14 cm, so the correct approach is:

Since object distance for eyepiece is u_e = L - v_o (distance between lenses minus image distance from objective).

Rearranged:
$$ L = v_o + u_e $$

Using lens formula for eyepiece:
$$ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \implies \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{2.5} = -0.04 - 0.4 = -0.44 $$

$$ u_e = -2.27 \text{ cm} $$

Therefore,
$$ L = v_o + u_e = 7.14 - 2.27 = 4.87 \text{ cm} $$

Magnifying power of microscope:
$$ M = m_o \times m_e = \left| \frac{v_o}{u_o} \right| \times \left( 1 + \frac{D}{f_e} \right) $$

Calculate:
$$ m_o = \frac{7.14}{0.9} = 7.93 $$

$$ m_e = 1 + \frac{25}{2.5} = 1 + 10 = 11 $$

$$ M = 7.93 \times 11 = 87.23 $$

Therefore, the separation between lenses is approximately 4.87 cm and magnifying power is about 87. — Step 1: Use lens formula for objective to find image distance.
Step 2: Use lens formula for eyepiece with final image at near point.
Step 3: Calculate separation between lenses.
Step 4: Calculate magnifying power as product of objective and eyepiece magnifications.

Question 13

13 A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Accepted Answer

Given:
Focal length of objective, f_o = 144 cm
Focal length of eyepiece, f_e = 6.0 cm

Magnifying power (angular magnification) of telescope for normal adjustment (final image at infinity):
$$ M = \frac{f_o}{f_e} = \frac{144}{6} = 24 $$

Separation between objective and eyepiece:
$$ L = f_o + f_e = 144 + 6 = 150 \text{ cm} $$

Therefore, magnifying power is 24 and separation between lenses is 150 cm. — Step 1: Use formula for magnifying power of telescope.
Step 2: Calculate separation as sum of focal lengths for normal adjustment.
Step 3: Substitute given values and compute.

Question 14

14 (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10^8 m, and the radius of lunar orbit is 3.8 × 10^8 m.

Accepted Answer

(a) Given:
Focal length of objective, f_o = 15 m = 1500 cm
Focal length of eyepiece, f_e = 1.0 cm

Angular magnification:
$$ M = \frac{f_o}{f_e} = \frac{1500}{1} = 1500 $$

(b) Diameter of moon, D_m = 3.48 × 10^6 m (Note: The question states 3.48 × 10^8 m, but actual diameter is ~3.48 × 10^6 m; assuming typo, use given value)
Radius of lunar orbit, R = 3.8 × 10^8 m

Diameter of image formed by objective lens:
$$ d_i = \frac{f_o}{R} \times D_m = \frac{15}{3.8 \times 10^8} \times 3.48 \times 10^8 = 15 \times \frac{3.48}{3.8} = 15 \times 0.9158 = 13.74 \text{ m} $$

Therefore, the image diameter is approximately 13.74 m. — Step 1: Calculate angular magnification using focal lengths.
Step 2: Use magnification formula for image size formed by objective lens.
Step 3: Substitute given values and calculate image diameter.

Question 15

15 Use the mirror equation to deduce that:

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Accepted Answer

(a) For concave mirror, focal length f > 0. Object distance u between f and 2f: -2f < u < -f Using mirror formula: $$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} $$ Since u is negative, 1/u is negative. Calculate v: For u = -1.5f (between f and 2f), $$ \frac{1}{v} = \frac{1}{f} - \left(-\frac{1}{1.5f}\right) = \frac{1}{f} + \frac{2}{3f} = \frac{5}{3f} \implies v = \frac{3f}{5} $$ But this is positive and greater than 2f, so image is real and beyond 2f. (b) For convex mirror, focal length f < 0. Using mirror formula: $$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} $$ Since f < 0 and u < 0, 1/f is negative and 1/u is negative. Therefore, 1/v is negative, so v is negative. Negative v means image is virtual and behind the mirror. (c) Magnification m = -v/u. Since v and u are negative, m is positive but less than 1 (diminished). Image lies between focus and pole. (d) For object between pole and focus (0 > u > -f), Using mirror formula, v is positive and greater than f. Magnification m = -v/u is negative and greater than 1 in magnitude. Therefore, image is virtual, enlarged, and on the same side as object. — Step 1: Use mirror formula and sign conventions. Step 2: Substitute object distances for each case. Step 3: Analyze sign and magnitude of image distance and magnification. Step 4: Deduce nature, size, and position of image algebraically.

Ray Optics And Optical Instruments 9.1 Introduction

Ray Optics And Optical Instruments 9.1 Introduction — Study Notes

9.1 Introduction

Practice Questions — Ray Optics And Optical Instruments 9.1 Introduction

All 6 Chapters in Physics Part-II