Class 12 Physics Notes – Chapter 9: Ray Optics and Optical Instruments

1. Reflection of Light by Spherical Mirrors

Light travels in a straight line, but when it strikes a reflecting surface, it bounces back. The laws of reflection state that the angle of incidence is equal to the angle of reflection, and the incident ray, reflected ray, and the normal all lie in the same plane.

Sign Convention (Cartesian)

• All distances are measured from the pole (P) of the mirror.
• Distances measured in the direction of incident light are positive (+), and opposite to it are negative (-).
• Heights measured upwards are positive (+), and downwards are negative (-).

Focal Length and Mirror Equation

The relation between the focal length (f) and the radius of curvature (R) of a spherical mirror is: f = R / 2.

Mirror Equation: 1/v + 1/u = 1/f

Linear Magnification (m): m = h’ / h = – v / u

Where u is object distance, v is image distance, h is object height, and h’ is image height.

2. Refraction

When light passes obliquely from one transparent medium to another, it bends. This is called refraction.

Snell’s Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.

Formula: (sin i) / (sin r) = n₂₁ = n₂ / n₁

Where n₂₁ is the refractive index of medium 2 with respect to medium 1.

3. Total Internal Reflection (TIR)

When light travels from an optically denser medium to a rarer medium at an angle of incidence greater than the critical angle (i_c), it is totally reflected back into the denser medium.

Critical Angle Formula: sin i_c = n₂ / n₁

Applications of TIR: Optical fibres (used in telecommunications and medical endoscopy), Mirages, and totally reflecting prisms.

4. Refraction at Spherical Surfaces and by Lenses

For refraction at a single spherical surface separating two media of refractive indices n₁ and n₂:

n₂ / v – n₁ / u = (n₂ – n₁) / R

Lens Maker’s Formula

It is used to design lenses of desired focal length:

1/f = (n₂₁ – 1) [ (1/R₁) – (1/R₂) ]

Thin Lens Formula & Power

Lens Formula: 1/v – 1/u = 1/f

Power of a Lens (P): It is the reciprocal of focal length (in metres). P = 1 / f. Its SI unit is Dioptre (D).

Combination of Thin Lenses: 1/f = 1/f₁ + 1/f₂ + 1/f₃ + …

5. Refraction through a Prism

When light passes through a triangular glass prism, it suffers deviation. The angle of deviation (δ) is given by:

δ = i + e – A (where A is the angle of the prism)

At minimum deviation (D_m), i = e and the refracted ray is parallel to the base of the prism. The refractive index is:

n₂₁ = sin [ (A + D_m) / 2 ] / sin(A / 2)

6. Optical Instruments

• Simple Microscope: A magnifying glass. Magnification m = 1 + (D/f) for image at near point, and m = D/f for image at infinity.
• Compound Microscope: Uses an objective and an eyepiece. Total magnification m = m_o × m_e.
• Telescope: Used for viewing distant objects. Magnification m = f_o / f_e (where f_o is focal length of objective and f_e is focal length of eyepiece).

Multiple Choice Questions (MCQs)

1. The relation between the focal length (f) and the radius of curvature (R) of a spherical mirror is:
a) f = R
b) f = 2R
c) f = R/2
d) f = R/4
Answer: (c) f = R/2

2. According to Snell’s Law, the refractive index n₂₁ is equal to:
a) sin r / sin i
b) sin i / sin r
c) cos i / cos r
d) tan i / tan r
Answer: (b) sin i / sin r

3. Total Internal Reflection (TIR) occurs when light travels from:
a) Rarer to denser medium
b) Denser to rarer medium
c) Vacuum to glass
d) Air to water
Answer: (b) Denser to rarer medium

4. Optical fibres are based on the principle of:
a) Diffraction
b) Dispersion
c) Interference
d) Total Internal Reflection
Answer: (d) Total Internal Reflection

5. The SI unit of power of a lens is:
a) Metre
b) Watt
c) Dioptre (D)
d) Joule
Answer: (c) Dioptre (D)

6. The thin lens formula is given by:
a) 1/v + 1/u = 1/f
b) 1/u – 1/v = 1/f
c) 1/v – 1/u = 1/f
d) v + u = f
Answer: (c) 1/v – 1/u = 1/f

7. When a prism is in the position of minimum deviation, the angle of incidence (i) and angle of emergence (e) are related as:
a) i > e
b) i < e
c) i = e
d) i + e = 90°
Answer: (c) i = e

8. Two thin lenses of focal lengths f₁ and f₂ are placed in contact. The equivalent focal length F is given by:
a) F = f₁ + f₂
b) 1/F = 1/f₁ + 1/f₂
c) F = (f₁ × f₂) / (f₁ – f₂)
d) F = f₁ / f₂
Answer: (b) 1/F = 1/f₁ + 1/f₂

9. In a compound microscope, the focal length of the objective lens is:
a) Equal to eyepiece
b) Much larger than eyepiece
c) Very small
d) Infinity
Answer: (c) Very small

10. The phenomenon of splitting of white light into its constituent colours is called:
a) Reflection
b) Dispersion
c) Interference
d) Polarisation
Answer: (b) Dispersion

Textbook Exercise Solutions

Q 9.1: 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.
Solution:
Given: Object size h = +2.5 cm, Object distance u = -27 cm. Radius of curvature R = -36 cm.
Focal length f = R/2 = -36/2 = -18 cm.
Using Mirror Equation: 1/v + 1/u = 1/f
1/v = 1/f – 1/u = 1/(-18) – 1/(-27) = (-3 + 2) / 54 = -1 / 54
So, v = -54 cm. The screen should be placed 54 cm in front of the mirror.
Magnification m = -v/u = -(-54) / (-27) = -2.
Image size h’ = m × h = -2 × 2.5 = -5.0 cm.
The image is real, inverted, and magnified.

Q 9.2: 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.
Solution:
Given: h = +4.5 cm, u = -12 cm, f = +15 cm (convex mirror).
Mirror Formula: 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.
The image is formed 6.67 cm behind the mirror (virtual).
Magnification m = -v/u = -(20/3) / (-12) = 20 / 36 = +0.55.
Size of image h’ = m × h = 0.55 × 4.5 = 2.5 cm.
As the needle is moved farther from the mirror, the image moves towards the focus and its size goes on decreasing.

Q 9.3: 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?
Solution:
Real depth = 12.5 cm, Apparent depth = 9.4 cm.
Refractive index of water n_w = Real depth / Apparent depth = 12.5 / 9.4 ≈ 1.33.
Now, new liquid has refractive index n_l = 1.63.
New Apparent depth = Real depth / n_l = 12.5 / 1.63 ≈ 7.67 cm.
Distance the microscope has to be moved = Initial apparent depth – Final apparent depth
Distance = 9.4 – 7.67 = 1.73 cm (upwards).

Q 9.5: A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. 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.
Solution:
Light will only emerge out from a circular area whose radius R corresponds to the critical angle (i_c).
Depth h = 80 cm = 0.8 m. Refractive index n = 1.33 = 4/3.
We know sin i_c = 1 / n = 1 / 1.33 = 0.75 ⇒ i_c ≈ 48.6°.
From geometry, R = h × tan(i_c) = 0.8 × tan(48.6°) = 0.8 × 1.134 = 0.9 m (approx).
Area A = π R² = 3.14 × (0.9)² = 3.14 × 0.81 ≈ 2.58 m².

Q 9.6: 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°.
Solution:
Given: Angle of prism A = 60°, Minimum deviation D_m = 40°.
Refractive index n = sin[(A+D_m)/2] / sin(A/2)
n = sin[(60+40)/2] / sin(60/2) = sin(50°) / sin(30°) = 0.766 / 0.5 = 1.532.

Q 9.7: 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 20cm?
Solution:
Given: n = 1.55, f = 20 cm. For a double convex lens of same radius, R₁ = +R and R₂ = -R.
Lens Maker’s Formula: 1/f = (n – 1) [1/R₁ – 1/R₂]
1/20 = (1.55 – 1) [1/R – (-1/R)]
1/20 = 0.55 × (2/R)
1/20 = 1.1 / R
R = 20 × 1.1 = 22 cm.

Q 9.8: A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
Solution:
The point P acts as a virtual object for the lens. So, u = +12 cm.
(a) For Convex Lens: f = +20 cm.
1/v – 1/u = 1/f ⇒ 1/v = 1/f + 1/u = 1/20 + 1/12 = (3 + 5) / 60 = 8 / 60
v = 60 / 8 = +7.5 cm. (Image forms 7.5 cm behind the lens).
(b) For Concave Lens: f = -16 cm.
1/v = 1/f + 1/u = 1/(-16) + 1/12 = (-3 + 4) / 48 = 1 / 48
v = +48 cm. (Image forms 48 cm behind the lens).