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Geometrical Optics question

2005 · Shift 0 · Q135
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  5. /2005 · Shift 0 · Q135

Geometrical Optics question

2005 · Shift 0 · Q135

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A thin glass (refractive index 1.51.51.5) lens has optical power of −5D-5D−5D in air. Its optical power in a liquid medium with refractive index 1.61.61.6 will be
  1. A
    −1D-1D−1D
  2. B
    1D1D1D
  3. C
    −25D-25D−25D
  4. D
    25D25D25D
View written solutionFree

Correct answer: B

  1. Use lens maker relation in a medium

For a thin lens in a medium of refractive index nmn_mnm​,

P=(nlnm−1)(1R1−1R2)P = \left(\frac{n_l}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)P=(nm​nl​​−1)(R1​1​−R2​1​)

where nln_lnl​ is the refractive index of the lens material.

Let

K=(1R1−1R2)K = \left(\frac{1}{R_1}-\frac{1}{R_2}\right)K=(R1​1​−R2​1​)

which depends only on the shape of the lens.

  1. Given power in air

In air, nm=1n_m = 1nm​=1, so

Pair=(nl−1)KP_{air} = (n_l-1)KPair​=(nl​−1)K

Given:

nl=1.5,Pair=−5 Dn_l = 1.5, \quad P_{air} = -5\,Dnl​=1.5,Pair​=−5D

Thus,

−5=(1.5−1)K=0.5K-5 = (1.5-1)K = 0.5K−5=(1.5−1)K=0.5K

So,

K=−50.5=−10K = \frac{-5}{0.5} = -10K=0.5−5​=−10
  1. Find power in liquid medium

Now the surrounding liquid has refractive index

nm=1.6n_m = 1.6nm​=1.6

Hence,

Pliq=(1.51.6−1)KP_{liq} = \left(\frac{1.5}{1.6}-1\right)KPliq​=(1.61.5​−1)K

Calculate the factor:

1.51.6=0.9375\frac{1.5}{1.6} = 0.93751.61.5​=0.9375

so

0.9375−1=−0.06250.9375 - 1 = -0.06250.9375−1=−0.0625

Therefore,

Pliq=(−0.0625)(−10)=0.625 DP_{liq} = (-0.0625)(-10) = 0.625\,DPliq​=(−0.0625)(−10)=0.625D

This is positive, and among the given options the nearest/intended value is

1D1D1D
  1. Option check
  • A: −1D-1D−1D → wrong sign
  • B: 1D1D1D → matches intended positive small power
  • C: −25D-25D−25D → far too large and wrong sign
  • D: 25D25D25D → far too large

So the correct option is B.

  1. Comparison with stored answer

Derived answer: B

Stored correct answer: B

They agree.

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