One plano-convex and one plano-concave lens of same radius of curvature 'R' but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is \mu 1 and that of 2 is \mu 2, then the focal length of the combination is :

R \over 2 - ( \mu _1 - \mu _2 )

Geometrical Optics PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q58 with Solution

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Correct answer: C

Interpret the arrangement

A plano-convex lens and a plano-concave lens are joined side by side as shown. Since they are side by side, each occupies half the aperture, but both have the same optical axis and the same curved radius magnitude $R$ .

For paraxial rays, the focal length of such a combination is determined by the power contributed by the refracting surfaces.

Use lens maker idea surface-wise

The outer plane surfaces have zero power, because for a plane surface:

P=\frac{\mu_2-\mu_1}{R}=0 \quad (R=\infty)

So only the two curved surfaces matter.

Power of the plano-convex part (material 1)

For a thin plano-convex lens in air, with curved surface radius $R$ :

P_1=(\mu_1-1)\left(\frac{1}{R}-0\right)=\frac{\mu_1-1}{R}

So its focal length is

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

Power of the plano-concave part (material 2)

For a thin plano-concave lens in air, the power is negative:

P_2=(\mu_2-1)\left(0-\frac{1}{R}\right)=-\frac{\mu_2-1}{R}

So its focal length is

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

Net power of the combination

Since the two parts are joined and act together for the incident beam, the resultant power is the algebraic sum:

P=P_1+P_2

Thus,

P=\frac{\mu_1-1}{R}-\frac{\mu_2-1}{R}

P=\frac{\mu_1-\mu_2}{R}

Hence the focal length $f$ is

\frac{1}{f}=\frac{\mu_1-\mu_2}{R}

So,

f=\frac{R}{\mu_1-\mu_2}

Match with options

f=\frac{R}{\mu_1-\mu_2}

This matches Option C.

Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

So they agree.

More from Geometrical Optics

View written solutionFree

Correct answer: C

Interpret the arrangement

For paraxial rays, the focal length of such a combination is determined by the power contributed by the refracting surfaces.

Use lens maker idea surface-wise

The outer plane surfaces have zero power, because for a plane surface:

P=\frac{\mu_2-\mu_1}{R}=0 \quad (R=\infty)

So only the two curved surfaces matter.

Power of the plano-convex part (material 1)

For a thin plano-convex lens in air, with curved surface radius $R$ :

P_1=(\mu_1-1)\left(\frac{1}{R}-0\right)=\frac{\mu_1-1}{R}

So its focal length is

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

Power of the plano-concave part (material 2)

For a thin plano-concave lens in air, the power is negative:

P_2=(\mu_2-1)\left(0-\frac{1}{R}\right)=-\frac{\mu_2-1}{R}

So its focal length is

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

Net power of the combination

Since the two parts are joined and act together for the incident beam, the resultant power is the algebraic sum:

P=P_1+P_2

Thus,

P=\frac{\mu_1-1}{R}-\frac{\mu_2-1}{R}

P=\frac{\mu_1-\mu_2}{R}

Hence the focal length $f$ is

\frac{1}{f}=\frac{\mu_1-\mu_2}{R}

So,

f=\frac{R}{\mu_1-\mu_2}

Match with options

f=\frac{R}{\mu_1-\mu_2}

This matches Option C.

Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

So they agree.

Geometrical Optics question

Geometrical Optics question

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