Ray Optics,, 2 Marker Questions

CBSE Class 12 Physics: Ray Optics 2-Mark PYQs

Interactive Topic-Wise Questions with Topper-Style Solutions

1. Reflection and Mirrors

Q1 (Delhi 2010): Object Placement for Concave Mirror

A concave mirror of focal length 20 cm forms an image at 40 cm. Where is the object placed?

Solution: Given: \( f = -20 \, \text{cm} \) (concave mirror), \( v = -40 \, \text{cm} \) (real image). Using mirror formula:

1/u + 1/v = 1/f

1/u + 1/(-40) = 1/(-20)

1/u = -1/20 + 1/40 = -1/40

u = -40 \, \text{cm}

Final Answer: Object is 40 cm in front of the mirror.

Q2 (All India 2011): Convex Mirror Object Distance

A convex mirror of focal length 30 cm forms an image half the size of the object. Find the object distance.

Solution: Given: \( f = +30 \, \text{cm} \), magnification \( m = +1/2 \). Magnification: \( m = -v/u \implies +1/2 = -v/u \implies v = -u/2 \). Mirror formula:

1/u + 1/v = 1/f

1/u + 1/(-u/2) = 1/30

1/u - 2/u = 1/30 \implies -1/u = 1/30 \implies u = -30 \, \text{cm}

Final Answer: Object is 30 cm in front of the mirror.

Q3: Plane vs. Convex Mirror Images

State two differences between images formed by a plane mirror and a convex mirror.

Solution:
1. Plane mirror: Image is virtual, same size (\( m = 1 \)). Convex mirror: Image is virtual, smaller (\( |m| < 1 \)).
2. Plane mirror: Image at same distance behind mirror. Convex mirror: Image between focus and mirror.
Final Answer: As above.

Q4: Focal Length of Concave Mirror

A concave mirror produces a real image at 30 cm when the object is at 15 cm. Calculate the focal length.

Solution: Given: \( u = -15 \, \text{cm} \), \( v = -30 \, \text{cm} \). Mirror formula:

1/u + 1/v = 1/f

1/(-15) + 1/(-30) = 1/f

-1/15 - 1/30 = -3/30 = 1/f

f = -10 \, \text{cm}

Final Answer: Focal length is 10 cm.

2. Refraction and Lenses

Q5 (Delhi 2008): Lens Disappearance

A glass lens of refractive index 1.5 is placed in a liquid. What must be the liquid’s refractive index to make the lens disappear?

Solution: For no refraction, the liquid’s refractive index (\( \mu_l \)) must equal the lens’s (\( \mu_g = 1.5 \)). Final Answer: Refractive index of liquid is 1.5.

Q6 (Delhi 2008): Lens in Same Medium

A converging lens of refractive index 1.5 is in a liquid of the same refractive index. What is its focal length?

Solution: Lens Maker’s formula: \( 1/f = (\mu_g/\mu_m - 1)(1/R_1 - 1/R_2) \). Given: \( \mu_g = 1.5 \), \( \mu_m = 1.5 \).

\mu_g/\mu_m = 1.5/1.5 = 1 \implies 1/f = (1 - 1)(1/R_1 - 1/R_2) = 0

Final Answer: Focal length is infinite.

Q7 (Delhi 2008): Lens Power Variation

How does the power of a convex lens vary if red light is replaced by violet light?

Solution: Power \( P = 1/f \), where \( 1/f = (\mu - 1)(1/R_1 - 1/R_2) \). Since \( \mu_{\text{violet}} > \mu_{\text{red}} \), focal length decreases, so power increases. Final Answer: Power increases.

Q8 (All India 2009): Lens Combination

Two thin lenses of power +4D and –2D are in contact. What is the focal length?

Solution: Power: \( P = P_1 + P_2 = +4 \, \text{D} + (-2 \, \text{D}) = +2 \, \text{D} \). Focal length:

f = 1/P = 1/2 = 0.5 \, \text{m} = 50 \, \text{cm}

Final Answer: Focal length is 50 cm.

3. Prisms

Q9 (All India 2008): Prism Deviation with Light Color

How does the angle of minimum deviation of a prism vary if violet light is replaced with red light?

Solution: Prism formula: \( \mu = \sin((A + \delta_m)/2)/\sin(A/2) \). Since \( \mu_{\text{violet}} > \mu_{\text{red}} \), lower \( \mu \) for red light reduces \( \delta_m \). Final Answer: Angle of minimum deviation decreases.

Q10 (All India 2008): Prism in Liquid

How does the angle of minimum deviation of a prism (μ = 1.5) change in a liquid (μ = 1.3)?

Solution: Effective \( \mu = 1.5/1.3 \approx 1.15 \). Prism formula: \( \mu = \sin((A + \delta_m)/2)/\sin(A/2) \). Lower \( \mu \) reduces \( \delta_m \). Final Answer: Angle of minimum deviation decreases.

4. Scattering of Light

Q11 (All India 2008): Sky Color

Why does the bluish colour predominate in a clear sky?

Solution: Rayleigh’s law: \( I \propto 1/\lambda^4 \). Blue light (shorter wavelength) scatters more than red light, dominating the sky’s color. Final Answer: Blue light scatters more due to \( 1/\lambda^4 \).

5. Optical Instruments

Q12 (Delhi 2009): Compound Microscope Diagram

Draw the labelled ray diagram for image formation by a compound microscope.

Solution: Object beyond objective’s focal point forms a real, inverted, magnified image. Eyepiece forms a virtual, enlarged image at the near point. Magnification: \( m = (v_o/u_o) \times (1 + D/f_e) \). [Diagram: Label object, objective lens, intermediate image, eyepiece, final image.] Final Answer: Diagram as described.

Q13 (All India 2009): Reflecting vs. Refracting Telescope

Write three advantages of a reflecting telescope over a refracting telescope.

Solution:
1. No chromatic aberration (mirrors don’t disperse light).
2. Larger apertures improve resolution.
3. Parabolic mirrors reduce spherical aberration.
Final Answer: As listed.

Practice Tips

- Reflection and Mirrors: Master mirror formula and sign conventions.
- Refraction and Lenses: Practice Lens Maker’s formula and power calculations.
- Prisms: Understand prism formula and refractive index effects.
- Scattering: Know Rayleigh’s law.
- Optical Instruments: Practice ray diagrams and telescope advantages.

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