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Geometrical Optics question

2025 · 29 Jan · Shift 2 · Q59
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  5. /2025 · 29 Jan · Shift 2 · Q59

Geometrical Optics question

2025 · 29 Jan · Shift 2 · Q59

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to
  1. A
    96 cm
  2. B
    72 cm
  3. C
    24 cm
  4. D
    48 cm
View written solutionFree

Correct answer: A

  1. Use the lens maker relation in a medium

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

1f=(nlnm−1)(1R1−1R2)\frac{1}{f}=\left(\frac{n_l}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(nm​nl​​−1)(R1​1​−R2​1​)

where:

  • nln_lnl​ = refractive index of lens material
  • nmn_mnm​ = refractive index of surrounding medium
  1. For the same lens, curvature term remains constant

Let

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

Then in air:

1fair=(nl−1)K\frac{1}{f_{air}}=(n_l-1)Kfair​1​=(nl​−1)K

since nm=1n_m=1nm​=1 for air.

Given:

nl=1.5,fair=24 cmn_l=1.5,\qquad f_{air}=24\text{ cm}nl​=1.5,fair​=24 cm

So,

124=(1.5−1)K=0.5K\frac{1}{24}=(1.5-1)K=0.5K241​=(1.5−1)K=0.5K

Hence,

K=112K=\frac{1}{12}K=121​
  1. Now immerse the lens in water

For water,

nm=1.33n_m=1.33nm​=1.33

So,

1fwater=(1.51.33−1)K\frac{1}{f_{water}}=\left(\frac{1.5}{1.33}-1\right)Kfwater​1​=(1.331.5​−1)K

First compute:

1.51.33≈1.1278\frac{1.5}{1.33}\approx 1.12781.331.5​≈1.1278

Thus,

1.51.33−1≈0.1278\frac{1.5}{1.33}-1\approx 0.12781.331.5​−1≈0.1278

Therefore,

1fwater=0.1278×112\frac{1}{f_{water}}=0.1278\times \frac{1}{12}fwater​1​=0.1278×121​ 1fwater≈0.01065\frac{1}{f_{water}}\approx 0.01065fwater​1​≈0.01065

So,

fwater≈93.9 cmf_{water}\approx 93.9\text{ cm}fwater​≈93.9 cm

which is closest to

96 cm96\text{ cm}96 cm
  1. Alternative ratio method

Using

fwaterfair=(nl−1)(nlnm−1)\frac{f_{water}}{f_{air}}=\frac{(n_l-1)}{\left(\frac{n_l}{n_m}-1\right)}fair​fwater​​=(nm​nl​​−1)(nl​−1)​ fwater=24×0.5(1.51.33−1)f_{water}=24\times \frac{0.5}{\left(\frac{1.5}{1.33}-1\right)}fwater​=24×(1.331.5​−1)0.5​ fwater≈24×0.50.1278≈24×3.91≈93.9 cmf_{water}\approx 24\times \frac{0.5}{0.1278}\approx 24\times 3.91\approx 93.9\text{ cm}fwater​≈24×0.12780.5​≈24×3.91≈93.9 cm

Again, nearest option is 969696 cm.

  1. Evaluate options
  • A: 969696 cm — Correct
  • B: 727272 cm — Incorrect
  • C: 242424 cm — Incorrect
  • D: 484848 cm — Incorrect

Therefore, the correct answer is Option A.

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