Combination of Lenses and Mirrors with Liquid Interfaces: Determining the equivalent focal length and final image position for systems involving glass lenses with liquid layers between them and a concave/convex mirror.

Combination of Lenses and Mirrors with Liquid Interfaces

Basic Concept

In a lens-liquid-mirror system, light may undergo refraction, reflection and again refraction. Therefore the complete ray path must be considered.

Lens in medium:

1/f = (μ_lens/μ_medium − 1) (1/R₁ − 1/R₂)

Power of lens:
P = 1/f

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

Spherical mirror:
f = R/2

Snell's Law:
μ₁ sin i = μ₂ sin r

Q1. A convex glass lens is immersed in a liquid whose refractive index is equal to glass. Its power becomes:
A. Maximum
B. Double
C. Nearly zero
D. Infinite

Answer: C

When μ_glass = μ_liquid, relative refractive index becomes 1. Hence lens power tends to zero.

Q2. A concave mirror has radius 40 cm. Its focal length is:
A. 10 cm
B. 20 cm
C. 40 cm
D. 80 cm

Answer: B

f = R/2 = 40/2 = 20 cm.

Q3. A lens has focal length 50 cm. Its power is:
A. 0.5 D
B. 1 D
C. 2 D
D. 5 D

Answer: C

f = 0.50 m
P = 1/f = 1/0.50 = +2 D.

Q4. If refractive index of liquid approaches that of glass, the focal length of a convex lens:
A. Decreases
B. Increases
C. Becomes zero
D. Remains fixed

Answer: B

Lens power decreases, therefore focal length increases.

Q5. For a concave mirror of R = 60 cm, focal length is:
A. +30 cm
B. −30 cm
C. +60 cm
D. −60 cm

Answer: B

For concave mirror under Cartesian convention: f = −R/2 = −30 cm.

Q6. Snell's law is:
A. μ₁ cos i = μ₂ cos r
B. μ₁ sin i = μ₂ sin r
C. μ₁ tan i = μ₂ tan r
D. i = 2r

Answer: B

μ₁ sin i = μ₂ sin r.

Q7. A convex lens of focal length 25 cm has power:
A. +2 D
B. +4 D
C. −4 D
D. +25 D

Answer: B

P = 1/0.25 = +4 D.

Q8. A concave mirror forms a real image when object is:
A. Between P and F
B. Beyond F
C. Behind mirror
D. At P

Answer: B

For object beyond focus, reflected rays converge and form a real image.

Q9. Increasing liquid refractive index toward glass causes lens power to:
A. Increase
B. Decrease
C. Become infinite
D. Remain same

Answer: B

Relative refractive index decreases, so power decreases.

Q10. Two lenses of +3 D and +2 D are in contact. Equivalent power:
A. +1 D
B. +5 D
C. −5 D
D. +6 D

Answer: B

P = P₁ + P₂ = 3 + 2 = +5 D.

Q11. Equivalent focal length for Q10 is:
A. 10 cm
B. 20 cm
C. 25 cm
D. 50 cm

Answer: B

f = 1/5 m = 0.20 m = 20 cm.

Q12. Focal length of a plane mirror is:
A. Zero
B. 1 m
C. Infinite
D. −1 m

Answer: C

For plane mirror R → ∞, therefore f → ∞.

Q13. A liquid layer between lens and mirror mainly affects:
A. Ray path
B. Mass
C. Frequency
D. Source speed

Answer: A

Liquid changes the refractive environment and hence ray direction.

Q14. Object at centre of curvature of concave mirror produces image:
A. At F
B. At C
C. Behind mirror
D. Infinity

Answer: B

Object at C gives image at C, same size and inverted.

Q15. A convex lens can behave as a diverging lens when:
A. μ_medium > μ_lens
B. μ_medium = 1
C. μ_lens is very high
D. None

Answer: A

If surrounding medium has higher refractive index, sign of lens power can reverse.

Q16. A lens has power −2 D. Its focal length is:
A. +0.5 m
B. −0.5 m
C. +2 m
D. −2 m

Answer: B

f = 1/P = −0.5 m.

Q17. If f = −15 cm for concave mirror, R is:
A. −7.5 cm
B. −15 cm
C. −30 cm
D. +30 cm

Answer: C

R = 2f = −30 cm.

Q18. At normal incidence, angle of refraction is:
A. 90°
B. 45°
C. 0°
D. 180°

Answer: C

At normal incidence i = 0°, hence r = 0°.

Q19. Two positive lenses in contact have combined power:
A. Difference
B. Sum
C. Zero
D. Product

Answer: B

P_eq = P₁ + P₂.

Q20. Object at infinity before a concave mirror forms image:
A. At P
B. At F
C. At C
D. Behind mirror

Answer: B

Parallel rays converge at principal focus.

Q21. Unit of optical power is:
A. Tesla
B. Dioptre
C. Watt
D. Pascal

Answer: B

Optical power is measured in dioptres.

Q22. Liquid changes lens power because it changes:
A. Relative refractive index
B. Mass
C. Radius of mirror
D. Object height

Answer: A

Lens power depends on relative refractive index.

Q23. Reflection law states:
A. i = r
B. i = 2r
C. i + r = 180°
D. i = 90°

Answer: A

Angle of incidence equals angle of reflection.

Q24. When liquid index approaches glass index, lens power:
A. Tends to zero
B. Tends to infinity
C. Doubles
D. Remains constant

Answer: A

Relative refractive index approaches unity.

Q25. A ray through optical centre of an ideal thin lens emerges:
A. Deviated
B. Undeviated
C. Reflected
D. Perpendicular

Answer: B

Central ray of an ideal thin lens is considered undeviated.

Q26. If focal length = 20 cm, power is:
A. 2 D
B. 5 D
C. 10 D
D. 20 D

Answer: B

f = 0.20 m
P = 1/0.20 = 5 D.

Q27. The best method for solving lens-mirror systems is:
A. Guessing
B. Sequential ray tracing
C. Ignoring liquid
D. Ignoring sign convention

Answer: B

Every optical element must be considered in sequence.

Q28. For a spherical mirror:
A. f = R
B. f = R/2
C. f = 2R
D. f = R²

Answer: B

The paraxial focal length is R/2.

Q29. A diverging lens has:
A. Positive power
B. Negative power
C. Infinite power
D. Always zero power

Answer: B

Diverging lens has negative focal length and negative power.

Q30. The first step in a complicated optical system should be:
A. Guess final image
B. Draw ray diagram
C. Ignore interface
D. Ignore mirror

Answer: B

A ray diagram clearly shows the complete sequence of refraction and reflection.

A: Both Assertion and Reason are true and Reason explains Assertion.
B: Both are true but Reason does not explain Assertion.
C: Assertion true, Reason false.
D: Assertion false, Reason true.

AR1. Assertion: A convex lens can lose its power in liquid.
Reason: Lens power depends on relative refractive index.

Answer: A

AR2. Assertion: A spherical mirror has f = R/2.
Reason: Focus lies midway between P and C.

Answer: A

AR3. Assertion: Plane mirror has infinite focal length.
Reason: Its radius of curvature is infinite.

Answer: A

AR4. Assertion: A convex lens can behave as a concave lens in a suitable medium.
Reason: Relative refractive index can change the sign of power.

Answer: A

AR5. Assertion: Liquid has no effect on lens.
Reason: Refraction does not depend on refractive index.

Answer: D

AR6. Assertion: Concave mirror can form real image.
Reason: Reflected rays can actually converge.

Answer: A

AR7. Assertion: Increasing liquid refractive index reduces convex lens power.
Reason: Relative refractive index decreases.

Answer: A

AR8. Assertion: Mirror focal length always changes in liquid.
Reason: f = R/2.

Answer: D

AR9. Assertion: Optical power is reciprocal of focal length.
Reason: P = 1/f.

Answer: A

AR10. Assertion: Sequential ray tracing is necessary.
Reason: Image from one element becomes object for next element.

Answer: A

1 Mark Questions

Q1. Define optical power.

P = 1/f, where f is in metre.

Q2. State relation between R and f.

f = R/2.

Q3. State Snell's law.

μ₁ sin i = μ₂ sin r.

Q4. What happens when μ_lens = μ_medium?

Ideal lens power becomes zero.

Q5. Focal length of plane mirror?

Infinite.

2 Mark Questions

Q6. Why does convex lens become less powerful in liquid?

Because relative refractive index decreases, reducing lens power.

Q7. R = 50 cm for concave mirror. Find f.

f = −R/2 = −25 cm.

Q8. Lens f = 40 cm. Find power.

f = 0.40 m
P = 1/0.40 = +2.5 D.

Q9. Why sequential calculation is required?

Because image produced by one element becomes object for the next.

Q10. Write lens and mirror equations.

Lens: 1/f = 1/v − 1/u.
Mirror: 1/f = 1/v + 1/u.

3 Mark Questions

Q11. Convex lens f = 20 cm. Find power.

P = 1/0.20 = +5 D.

Q12. Explain effect of liquid refractive index.

Increasing liquid refractive index reduces relative refractive index, therefore decreases power and increases focal length.

Q13. Concave mirror f = −20 cm, u = −60 cm. Find v.

−1/20 = 1/v − 1/60
1/v = −1/30
v = −30 cm.

Q14. Find magnification for Q13.

m = −v/u = −0.5.

Q15. Why should liquid interface be considered?

Because refractive index change bends rays and changes image position.

4 Mark Questions

Q16. Explain change in focal length when liquid refractive index increases.

Relative refractive index decreases, lens power decreases and focal length increases.

Q17. Concave mirror f = −20 cm, object u = −40 cm. Find image.

Using mirror formula:
v = −40 cm.
Image is real, inverted and same size.

Q18. Two lenses +4 D and −1 D are in contact. Find equivalent focal length.

P = 4 − 1 = +3 D.
f = 1/3 m ≈ 33.3 cm.

Q19. Write procedure for solving lens-liquid-mirror system.

Calculate lens → find first image → use it as mirror object → find reflected image → trace back → use lens again → final image.

Q20. R = 80 cm for concave mirror. Find f.

f = −80/2 = −40 cm.

5 Mark Questions

Q21. Write lens-maker equation in surrounding medium.

1/f = (μ_lens/μ_medium − 1) (1/R₁ − 1/R₂).

Q22. Concave mirror f = −15 cm, u = −30 cm. Find v and m.

v = −30 cm.
m = −1.
Image is real, inverted and same size.

Q23. Explain why mirror must be included in final image calculation.

Mirror reverses ray direction and sends rays back through the liquid and lens. Therefore the lens acts again during return path.

Q24. Two lenses have powers +6 D and −2 D. Find f.

P = +4 D.
f = 1/4 = 0.25 m = 25 cm.

Q25. Explain Snell's law at liquid-glass interface.

μ₁ sin i = μ₂ sin r.
The ray bends toward or away from normal depending on refractive indices.

6 Mark Questions

Q26. Explain complete method for convex lens-liquid-concave mirror system.

1. Find effective focal length in liquid.
2. Find first image by lens.
3. Treat first image as mirror object.
4. Apply mirror equation.
5. Trace reflected rays through liquid.
6. Apply lens equation again to find final image.

Q27. Derive effect of surrounding medium on focal length.

1/f = (μ_l/μ_m − 1) (1/R₁ − 1/R₂).

As μ_m approaches μ_l, power tends to zero and focal length increases.

Q28. Give complete numerical strategy for lens-liquid-mirror system.

Draw diagram → choose sign convention → calculate effective lens → find first image → mirror calculation → reflected image → return through liquid → second lens calculation → final image.

Q29. Concave mirror f = −20 cm and u = −60 cm. Find v and m.

1/−20 = 1/v + 1/−60
1/v = −1/30
v = −30 cm.
m = −(−30)/(−60) = −0.5.

Q30. Discuss factors affecting final image.

Final image depends on lens index, liquid index, radii of lens, mirror radius, object distance, refraction, reflection and sequential ray propagation through the entire system.

1/f = (μ_lens/μ_medium − 1) (1/R₁ − 1/R₂)

P = 1/f

P_eq = P₁ + P₂ + P₃ + ...

1/f = 1/v + 1/u

f = R/2

m = −v/u

μ₁ sin i = μ₂ sin r

MCQ:

1-C, 2-B, 3-C, 4-B, 5-B, 6-B, 7-B, 8-B, 9-B, 10-B, 11-B, 12-C, 13-A, 14-B, 15-A, 16-B, 17-C, 18-C, 19-B, 20-B, 21-B, 22-A, 23-A, 24-A, 25-B, 26-B, 27-B, 28-B, 29-B, 30-B.

Assertion–Reason:

1-A, 2-A, 3-A, 4-A, 5-D, 6-A, 7-A, 8-D, 9-A, 10-A.