JEE Challenger
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PhyMedAdv 2014 • Paper 2

Focal Length of Thin Lens Combinations in Contact

Four distinct combinations of two thin lenses are presented in List I. Every curved surface has a radius of curvature equal to rr, and all lenses have a refractive index of 1.51.5. Pair each lens arrangement in List I with its effective focal length in List II, and select the correct option.

List IList IIP. [Lens combination P]1. 2rQ. [Lens combination Q]2. r/2R. [Lens combination R]3. −rS. [Lens combination S]4. r\begin{array}{ll} \text{List I} & \text{List II} \\ \text{P. [Lens combination P]} & 1.\ 2r \\ \text{Q. [Lens combination Q]} & 2.\ r/2 \\ \text{R. [Lens combination R]} & 3.\ -r \\ \text{S. [Lens combination S]} & 4.\ r \end{array}
Question Diagram 1
Question Diagram 2
Question Diagram 3
Question Diagram 4

Options

A

P-1, Q-2, R-3, S-4

B

P-2, Q-4, R-3, S-1

Correct
C

P-4, Q-1, R-2, S-3

D

P-2, Q-1, R-3, S-4

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and MirrorsReflection and Refraction

Step-by-Step Solution

To find the effective focal length of each combination of thin lenses, we use the Lens Maker's Formula:

1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

Given that the refractive index of all lenses is μ=1.5\mu = 1.5 and the magnitude of the radius of curvature of each curved surface is rr.

For thin lenses in contact, the effective focal length FF is given by:

1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}

1. Combination P:

Combination P consists of two identical equiconvex lenses in contact.

For a single equiconvex lens: 1f1=(1.5−1)(1r−(−1r))=0.5×2r=1r  ⟹  f1=r\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{r} - \left(-\frac{1}{r}\right)\right) = 0.5 \times \frac{2}{r} = \frac{1}{r} \implies f_1 = r

For two such lenses in contact: 1FP=1f1+1f1=1r+1r=2r  ⟹  FP=r2\frac{1}{F_P} = \frac{1}{f_1} + \frac{1}{f_1} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r} \implies F_P = \frac{r}{2}

Thus, P matches with 2.

2. Combination Q:

Combination Q consists of two identical plano-convex lenses in contact.

For a single plano-convex lens: 1f1=(1.5−1)(1r−1∞)=0.5r=12r  ⟹  f1=2r\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{r} - \frac{1}{\infty}\right) = \frac{0.5}{r} = \frac{1}{2r} \implies f_1 = 2r

For two such lenses in contact: 1FQ=12r+12r=1r  ⟹  FQ=r\frac{1}{F_Q} = \frac{1}{2r} + \frac{1}{2r} = \frac{1}{r} \implies F_Q = r

Thus, Q matches with 4.

3. Combination R:

Combination R consists of two identical plano-concave lenses in contact.

For a single plano-concave lens: 1f1=(1.5−1)(1∞−1r)=−0.5r=−12r  ⟹  f1=−2r\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{\infty} - \frac{1}{r}\right) = -\frac{0.5}{r} = -\frac{1}{2r} \implies f_1 = -2r

For two such lenses in contact: 1FR=(−12r)+(−12r)=−1r  ⟹  FR=−r\frac{1}{F_R} = \left(-\frac{1}{2r}\right) + \left(-\frac{1}{2r}\right) = -\frac{1}{r} \implies F_R = -r

Thus, R matches with 3.

4. Combination S:

Combination S consists of one plano-concave lens and one equiconvex lens in contact.

  • For the plano-concave lens: f1=−2rf_1 = -2r
  • For the equiconvex lens: f2=rf_2 = r

The effective focal length FSF_S is: 1FS=1f1+1f2=−12r+1r=12r  ⟹  FS=2r\frac{1}{F_S} = \frac{1}{f_1} + \frac{1}{f_2} = -\frac{1}{2r} + \frac{1}{r} = \frac{1}{2r} \implies F_S = 2r

Thus, S matches with 1.

Conclusion:

The correct matching is: P-2, Q-4, R-3, S-1\text{P-2, Q-4, R-3, S-1}

This corresponds to Option B.

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