JEE Challenger
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PhyHardAdv 2019 • Paper 1

Focal Length Variation in Thin Lens Made of Two Refractive Materials

A thin convex lens is constructed from two materials with refractive indices n1n_1 and n2n_2, as illustrated in the figure. The radii of curvature of both spherical surfaces are equal. Let ff be the focal length when n1=n2=nn_1 = n_2 = n. The focal length is f+Δff + \Delta f when n1=nn_1 = n and n2=n+Δnn_2 = n + \Delta n. Assuming Δ\Delta n \ll (n - 1) and 1<n<21 < n < 2, which of the following statement(s) is/are correct?

Question Diagram 1

Options

A

∣Δff∣<∣Δnn∣\left|\frac{\Delta f}{f}\right| < \left|\frac{\Delta n}{n}\right|

B

For n=1.5n = 1.5, Δn=10−3\Delta n = 10^{-3}, and f=20 cmf = 20\text{ cm}, the magnitude ∣Δf∣|\Delta f| equals 0.02 cm0.02\text{ cm} (rounded off to 2nd decimal place).

Correct
C

If Δnn<0\frac{\Delta n}{n} < 0, then Δff>0\frac{\Delta f}{f} > 0

Correct
D

The relationship between Δff\frac{\Delta f}{f} and Δnn\frac{\Delta n}{n} stays unchanged if both convex surfaces are replaced with concave surfaces of the same radius of curvature.

Correct

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and MirrorsReflection and Refraction

Step-by-Step Solution

To determine the correct statements, we analyze the focal length of the thin lens composed of two halves placed together.

1. Derivation of the Focal Length Relation

The lens is split vertically down the middle into two thin plano-convex lenses in contact.

For the left half made of material with refractive index n1n_1:

  • The radius of the left spherical surface is R1=+RR_1 = +R.
  • The flat surface has R2=∞R_2 = \infty.
  • The optical power of the left half is: P1=(n1−1)(1R−1∞)=n1−1RP_1 = (n_1 - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{n_1 - 1}{R}

For the right half made of material with refractive index n2n_2:

  • The flat surface has R1′=∞R_1' = \infty.
  • The radius of the right spherical surface is R2′=−RR_2' = -R.
  • The optical power of the right half is: P2=(n2−1)(1∞−(−1R))=n2−1RP_2 = (n_2 - 1)\left(\frac{1}{\infty} - \left(-\frac{1}{R}\right)\right) = \frac{n_2 - 1}{R}

The total power of the combination is: P=P1+P2=n1+n2−2RP = P_1 + P_2 = \frac{n_1 + n_2 - 2}{R}

Case 1: n1=n2=nn_1 = n_2 = n

Let ff be the focal length in this case: 1f=2(n−1)R  ⟹  R=2(n−1)f\frac{1}{f} = \frac{2(n - 1)}{R} \implies R = 2(n - 1)f

Case 2: n1=nn_1 = n and n2=n+Δnn_2 = n + \Delta n

The new focal length is f+Δff + \Delta f: 1f+Δf=n−1+n+Δn−1R=2(n−1)+ΔnR\frac{1}{f + \Delta f} = \frac{n - 1 + n + \Delta n - 1}{R} = \frac{2(n - 1) + \Delta n}{R}

Substitute R=2(n−1)fR = 2(n - 1)f into the equation: 1f+Δf=2(n−1)+Δn2(n−1)f=1f(1+Δn2(n−1))\frac{1}{f + \Delta f} = \frac{2(n - 1) + \Delta n}{2(n - 1)f} = \frac{1}{f} \left( 1 + \frac{\Delta n}{2(n - 1)} \right)

Taking the reciprocal using the binomial approximation (1+x)−1≈1−x(1 + x)^{-1} \approx 1 - x for x=Δn2(n−1)≪1x = \frac{\Delta n}{2(n - 1)} \ll 1: f+Δf≈f(1−Δn2(n−1))f + \Delta f \approx f \left( 1 - \frac{\Delta n}{2(n - 1)} \right)

Δf=−fΔn2(n−1)\Delta f = -\frac{f \Delta n}{2(n - 1)}

Dividing by ff, we get the fractional change in focal length: Δff=−Δn2(n−1)=−n2(n−1)(Δnn)\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)} = -\frac{n}{2(n - 1)} \left(\frac{\Delta n}{n}\right)

2. Evaluation of the Options

  • Option A: Taking the absolute value: ∣Δff∣=n2(n−1)∣Δnn∣\left|\frac{\Delta f}{f}\right| = \frac{n}{2(n - 1)} \left|\frac{\Delta n}{n}\right| Given 1<n<21 < n < 2, we have n−2<0  ⟹  2(n−1)<nn - 2 < 0 \implies 2(n - 1) < n. Thus, n2(n−1)>1\frac{n}{2(n - 1)} > 1, which implies: ∣Δff∣>∣Δnn∣\left|\frac{\Delta f}{f}\right| > \left|\frac{\Delta n}{n}\right| Hence, Option A is incorrect.

  • Option B: Given n=1.5n = 1.5, Δn=10−3\Delta n = 10^{-3}, and f=20 cmf = 20\text{ cm}: ∣Δf∣=fΔn2(n−1)=20×10−32(1.5−1)=20×10−31=0.02 cm|\Delta f| = \frac{f \Delta n}{2(n - 1)} = \frac{20 \times 10^{-3}}{2(1.5 - 1)} = \frac{20 \times 10^{-3}}{1} = 0.02\text{ cm} Hence, Option B is correct.

  • Option C: Since Δff=−n2(n−1)(Δnn)\frac{\Delta f}{f} = -\frac{n}{2(n - 1)} \left(\frac{\Delta n}{n}\right) and n2(n−1)>0\frac{n}{2(n - 1)} > 0 for 1<n<21 < n < 2: If Δnn<0\frac{\Delta n}{n} < 0, then Δff>0\frac{\Delta f}{f} > 0. Hence, Option C is correct.

  • Option D: If the convex surfaces are replaced by concave surfaces of radius RR:

    • Power of left half: P1=−n1−1RP_1 = -\frac{n_1 - 1}{R}
    • Power of right half: P2=−n2−1RP_2 = -\frac{n_2 - 1}{R}
    • Total power: P=−n1+n2−2RP = -\frac{n_1 + n_2 - 2}{R}

    For n1=n2=nn_1 = n_2 = n: 1f=−2(n−1)R  ⟹  R=−2(n−1)f\frac{1}{f} = -\frac{2(n - 1)}{R} \implies R = -2(n - 1)f

    For n1=nn_1 = n and n2=n+Δnn_2 = n + \Delta n: 1f+Δf=−2(n−1)+ΔnR=2(n−1)+Δn2(n−1)f=1f(1+Δn2(n−1))\frac{1}{f + \Delta f} = -\frac{2(n - 1) + \Delta n}{R} = \frac{2(n - 1) + \Delta n}{2(n - 1)f} = \frac{1}{f}\left( 1 + \frac{\Delta n}{2(n - 1)} \right)

    Applying binomial approximation gives the exact same relation: Δff=−Δn2(n−1)\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)} Hence, Option D is correct.

Conclusion

The correct options are B, C, and D.

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