JEE Challenger
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PhyMedAdv 2026 • Paper 1

Power Variation of Double Convex Lens Immersed in Liquid

A double convex lens made of glass of refractive index 1.51.5 and radii of curvature of the curved surfaces 20 cm20\text{ cm} each is immersed in a liquid of refractive index nLn_L. The correct plot showing the variation of the power, in the units of diopter (DD), as a function of nLn_L is:

Options

A
Option A
Correct
B
Option B
Correct
C
Option C
D
Option D

Topics & Concepts

Ray Optics and Optical InstrumentsLenses and MirrorsReflection and Refraction

Step-by-Step Solution

To determine the variation of the power of the double convex lens as a function of the refractive index of the liquid nLn_L, we use the geometry of the lens and optics principles.

1. Geometric Parameters of the Lens

For a double convex lens:

  • Refractive index of the lens, ng=1.5n_g = 1.5
  • Radius of curvature of the first surface, R1=+20 cm=+0.2 mR_1 = +20\text{ cm} = +0.2\text{ m}
  • Radius of curvature of the second surface, R2=−20 cm=−0.2 mR_2 = -20\text{ cm} = -0.2\text{ m}

The curvature factor is calculated as: (1R1−1R2)=10.2 m−(−10.2 m)=5 m−1+5 m−1=10 m−1\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{0.2\text{ m}} - \left( -\frac{1}{0.2\text{ m}} \right) = 5\text{ m}^{-1} + 5\text{ m}^{-1} = 10\text{ m}^{-1}

2. Definition 1: Power defined as P=1fP = \frac{1}{f}

Using the Lens Maker's Formula for a thin lens immersed in a medium of refractive index nLn_L: 1f=(ngnL−1)(1R1−1R2)\frac{1}{f} = \left( \frac{n_g}{n_L} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

Substituting the given values, the power PP in diopters (D\text{D}) is: P(nL)=(1.5nL−1)×10=15nL−10 DP(n_L) = \left( \frac{1.5}{n_L} - 1 \right) \times 10 = \frac{15}{n_L} - 10\text{ D}

Evaluating P(nL)P(n_L) at characteristic values of nLn_L:

  • For nL=1.0n_L = 1.0: P(1.0)=151.0−10=+5 DP(1.0) = \frac{15}{1.0} - 10 = +5\text{ D}
  • For nL=1.5n_L = 1.5: P(1.5)=151.5−10=0 DP(1.5) = \frac{15}{1.5} - 10 = 0\text{ D}
  • For nL=2.0n_L = 2.0: P(2.0)=152.0−10=−2.5 DP(2.0) = \frac{15}{2.0} - 10 = -2.5\text{ D}

Curve Characteristics:

  • First derivative: dPdnL=−15nL2<0\frac{dP}{dn_L} = -\frac{15}{n_L^2} < 0 (monotonically decreasing).
  • Second derivative: d2PdnL2=30nL3>0\frac{d^2P}{dn_L^2} = \frac{30}{n_L^3} > 0 (convex downwards).

This non-linear hyperbolic relationship corresponds to Option (A).

3. Definition 2: Power defined as the sum of surface refractivity P=(ng−nL)(1R1−1R2)P = (n_g - n_L)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

The optical power of refraction at two spherical boundaries is given by: P=ng−nLR1+nL−ngR2=(ng−nL)(1R1−1R2)P = \frac{n_g - n_L}{R_1} + \frac{n_L - n_g}{R_2} = (n_g - n_L)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

Substituting the given values: P(nL)=(1.5−nL)×10=15−10nL DP(n_L) = (1.5 - n_L) \times 10 = 15 - 10 n_L\text{ D}

Evaluating P(nL)P(n_L) at characteristic values of nLn_L:

  • For nL=1.0n_L = 1.0: P(1.0)=15−10(1.0)=+5 DP(1.0) = 15 - 10(1.0) = +5\text{ D}
  • For nL=1.5n_L = 1.5: P(1.5)=15−10(1.5)=0 DP(1.5) = 15 - 10(1.5) = 0\text{ D}
  • For nL=2.0n_L = 2.0: P(2.0)=15−10(2.0)=−5 DP(2.0) = 15 - 10(2.0) = -5\text{ D}

Curve Characteristics:

  • The relationship is linear with a constant negative slope of −10-10.

This linear relationship corresponds to Option (B).

Conclusion

Depending on the optical convention used for power of a lens in a medium, both plots (A) and (B) accurately represent the variation.

Correct Options: A or B

Power Variation of Double Convex Lens Immersed in Liquid | Physics PYQ Solution - JEE Challenger