A plano convex lens of refractive index \mu 1 and focal length ƒ1 is kept in contact with another plano concave lens of refractive index \mu 2 and focal length ƒ2. If the radius of curvature of their spherical faces is R each and ƒ1 = 2ƒ2, then \mu 1 and \mu 2 are related as -

Geometrical Optics PYQ — JEE Main 2019 · 10 Jan · Shift 1 · Q52 with Solution

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Correct answer: C

For a thin lens in air,

\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

We are given that both lenses have one plane face and one spherical face of radius of curvature $R$ .

For a plano-convex lens, one surface is plane and the other is convex.

Taking the plane surface radius as $\infty$ and the spherical surface radius as $R$ , its focal length magnitude is

\frac{1}{f_1}=(\mu_1-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{\mu_1-1}{R}

So,

f_1=\frac{R}{\mu_1-1}

For a plano-concave lens, one surface is plane and the other is concave.

Its power is negative, so

\frac{1}{f_2}=-(\mu_2-1)\frac{1}{R}

Hence the focal length magnitude is

|f_2|=\frac{R}{\mu_2-1}

Since the options involve only $\mu_1,\mu_2$ , clearly the question uses focal lengths by magnitude. Thus,

f_2=\frac{R}{\mu_2-1}

for comparison of magnitudes.

Substitute:

\frac{R}{\mu_1-1}=2\cdot \frac{R}{\mu_2-1}

Cancel $R$ :

\frac{1}{\mu_1-1}=\frac{2}{\mu_2-1}

Cross-multiply:

\mu_2-1=2(\mu_1-1)

\mu_2-1=2\mu_1-2

2\mu_1-\mu_2=1

This corresponds to:

\boxed{2\mu_1-\mu_2=1}

So the correct option is C.

Stored correct answer: C

My derived answer: C

They agree.