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Geometrical Optics question

2025 · 24 Jan · Shift 1 · Q64
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Geometrical Optics question

2025 · 24 Jan · Shift 1 · Q64

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m . The radius of curvature of the curved surface of the lens is
  1. A
    0.15 m
  2. B
    0.10 m
  3. C
    0.25 m
  4. D
    0.20 m
View written solutionFree

Correct answer: B

  1. Lens in a medium: relative refractive index

For a lens of refractive index ng=1.5n_g=1.5ng​=1.5 immersed in a liquid of refractive index nl=1.2n_l=1.2nl​=1.2, the effective refractive index is

μ=ngnl=1.51.2=1.25\mu = \frac{n_g}{n_l} = \frac{1.5}{1.2} = 1.25μ=nl​ng​​=1.21.5​=1.25
  1. Focal length of the plano-convex lens in the liquid

Using the lens maker formula in a medium:

1f=(μ−1)(1R1−1R2)\frac{1}{f} = \left(\mu-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(μ−1)(R1​1​−R2​1​)

For a plano-convex lens, one surface is plane and the other has radius RRR. Taking the first surface plane: R1=∞R_1=\inftyR1​=∞, second surface curved: R2=−RR_2=-RR2​=−R. So,

1f=(1.25−1)(0−(−1R))=0.25⋅1R\frac{1}{f} = (1.25-1)\left(0-\left(-\frac{1}{R}\right)\right) =0.25\cdot \frac{1}{R}f1​=(1.25−1)(0−(−R1​))=0.25⋅R1​

Hence,

f=4Rf = 4Rf=4R
  1. Effect of silvering the plane face

When the plane face is silvered, the light passes through the lens twice, with reflection at the plane mirror in between. The equivalent power becomes double the lens power (plane mirror has zero power), so the equivalent focal length FFF is

1F=2f⇒F=f2\frac{1}{F}=\frac{2}{f} \quad\Rightarrow\quad F=\frac{f}{2}F1​=f2​⇒F=2f​

Given that it behaves like a concave mirror of focal length 0.2 m0.2\,\text{m}0.2m,

F=0.2 mF=0.2\,\text{m}F=0.2m

Therefore,

f2=0.2⇒f=0.4 m\frac{f}{2}=0.2 \quad\Rightarrow\quad f=0.4\,\text{m}2f​=0.2⇒f=0.4m
  1. Find the radius of curvature

Since f=4Rf=4Rf=4R,

0.4=4R⇒R=0.1 m0.4=4R \quad\Rightarrow\quad R=0.1\,\text{m}0.4=4R⇒R=0.1m
  1. Option check

Thus the radius of curvature is

0.10 m\boxed{0.10\,\text{m}}0.10m​

So the correct option is B.

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