A planoconvex lens becomes an optical system of 28cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A. If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig-B, it acts like an optical system of focal length 10cm. The refractive index of the material of lens is :

Geometrical Optics PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q57 with Solution

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Correct answer: B

Let the focal length of the unsilvered plano-convex lens be $f$ .

For a plano-convex lens in air,

\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{\mu-1}{R}

f=\frac{R}{\mu-1}.

Case A: plane surface silvered

When the plane face is silvered, light first passes through the lens, reflects at the plane mirror, and again passes through the lens.

A plane mirror has zero optical power, so the lens is effectively traversed twice. Hence equivalent power is

P_A=2P=\frac{2}{f}

and equivalent focal length is

F_A=\frac{f}{2}.

Given

F_A=28\text{ cm}

therefore

f=56\text{ cm}.

So,

\frac{R}{\mu-1}=56 \qquad ...(1)

Case B: curved surface silvered

Now the curved face is silvered and light is incident from the plane side.

The system behaves as:

refraction at plane surface: no power,
reflection at curved silvered surface: acts like a spherical mirror inside medium,
refraction again at plane surface: again no power.

So the effective power comes only from the curved mirror.

For a spherical mirror, focal length in the medium is

f_m=\frac{R}{2}.

But the final image is observed in air through the plane surface. A plane refracting surface does not change vergence for paraxial rays except for refractive-index scaling, so the effective focal length in air becomes

F_B=\frac{R}{2\mu}.

Given

F_B=10\text{ cm}

thus

\frac{R}{2\mu}=10 \quad\Rightarrow\quad R=20\mu. \qquad ...(2)

Use (1) and (2)

From (1):

R=56(\mu-1)

From (2):

R=20\mu

Equating,

56(\mu-1)=20\mu

56\mu-56=20\mu

36\mu=56

\mu=\frac{56}{36}=\frac{14}{9}\approx 1.56

This is closest to

\boxed{1.55}

Option check

A: $1.50$ ❌
B: $1.55$ ✅
C: $1.75$ ❌
D: $1.51$ ❌

Therefore the correct option is

\boxed{\text{B }(1.55)}

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View written solutionFree

Correct answer: B

Let the focal length of the unsilvered plano-convex lens be $f$ .

For a plano-convex lens in air,

\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{\mu-1}{R}

f=\frac{R}{\mu-1}.

Case A: plane surface silvered

When the plane face is silvered, light first passes through the lens, reflects at the plane mirror, and again passes through the lens.

A plane mirror has zero optical power, so the lens is effectively traversed twice. Hence equivalent power is

P_A=2P=\frac{2}{f}

and equivalent focal length is

F_A=\frac{f}{2}.

Given

F_A=28\text{ cm}

therefore

f=56\text{ cm}.

So,

\frac{R}{\mu-1}=56 \qquad ...(1)

Case B: curved surface silvered

Now the curved face is silvered and light is incident from the plane side.

The system behaves as:

refraction at plane surface: no power,
reflection at curved silvered surface: acts like a spherical mirror inside medium,
refraction again at plane surface: again no power.

So the effective power comes only from the curved mirror.

For a spherical mirror, focal length in the medium is

f_m=\frac{R}{2}.

F_B=\frac{R}{2\mu}.

Given

F_B=10\text{ cm}

thus

\frac{R}{2\mu}=10 \quad\Rightarrow\quad R=20\mu. \qquad ...(2)

Use (1) and (2)

From (1):

R=56(\mu-1)

From (2):

R=20\mu

Equating,

56(\mu-1)=20\mu

56\mu-56=20\mu

36\mu=56

\mu=\frac{56}{36}=\frac{14}{9}\approx 1.56

This is closest to

\boxed{1.55}

Option check

A: $1.50$ ❌
B: $1.55$ ✅
C: $1.75$ ❌
D: $1.51$ ❌

Therefore the correct option is

\boxed{\text{B }(1.55)}

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