A plano-convex lens (focal length f2, refractive index \mu 2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index \mu 1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be :

Geometrical Optics PYQ — JEE Main 2019 · 12 Jan · Shift 2 · Q60 with Solution

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Correct answer: C

Identify the system

We have:

a plano-convex lens of refractive index $\mu_2$
a plano-concave lens of refractive index $\mu_1$
both have the same curved surface radius $R$
they fit exactly into each other, so the two curved surfaces coincide
the outer plane faces are parallel

This behaves like a single lens system with only one effective curved refracting surface contributing, because the outer faces are plane.

Use lens maker's formula for each lens

For a thin lens in air,

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

Plano-convex lens

Take its first surface plane and second surface curved. Then

R_1=\infty,\qquad R_2=-R

So,

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

Hence,

f_{pcx}=\frac{R}{\mu_2-1}

Plano-concave lens

Similarly,

R_1=\infty,\qquad R_2=+R

So,

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

Hence its focal length magnitude is

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

with negative sign for actual focal length. But we do not really need to use $f_1,f_2$ directly.

Equivalent power of the combination

Since the two lenses are in contact, powers add:

P=P_1+P_2

Now the plano-convex gives positive power,

P_2=\frac{\mu_2-1}{R}

The plano-concave gives negative power,

P_1=-\frac{\mu_1-1}{R}

Therefore,

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

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

So the equivalent focal length is

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

Match with options

Thus, the focal length of the combination is

\boxed{\frac{R}{\mu_2-\mu_1}}

which corresponds to Option C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

More from Geometrical Optics

View written solutionFree

Correct answer: C

Identify the system

We have:

a plano-convex lens of refractive index $\mu_2$
a plano-concave lens of refractive index $\mu_1$
both have the same curved surface radius $R$
they fit exactly into each other, so the two curved surfaces coincide
the outer plane faces are parallel

This behaves like a single lens system with only one effective curved refracting surface contributing, because the outer faces are plane.

Use lens maker's formula for each lens

For a thin lens in air,

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

Plano-convex lens

Take its first surface plane and second surface curved. Then

R_1=\infty,\qquad R_2=-R

So,

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

Hence,

f_{pcx}=\frac{R}{\mu_2-1}

Plano-concave lens

Similarly,

R_1=\infty,\qquad R_2=+R

So,

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

Hence its focal length magnitude is

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

with negative sign for actual focal length. But we do not really need to use $f_1,f_2$ directly.

Equivalent power of the combination

Since the two lenses are in contact, powers add:

P=P_1+P_2

Now the plano-convex gives positive power,

P_2=\frac{\mu_2-1}{R}

The plano-concave gives negative power,

P_1=-\frac{\mu_1-1}{R}

Therefore,

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

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

So the equivalent focal length is

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

Match with options

Thus, the focal length of the combination is

\boxed{\frac{R}{\mu_2-\mu_1}}

which corresponds to Option C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

Geometrical Optics question

Geometrical Optics question

Plano-convex lens

Plano-concave lens

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