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Geometrical Optics question

2025 · 24 Jan · Shift 1 · Q53
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  5. /2025 · 24 Jan · Shift 1 · Q53

Geometrical Optics question

2025 · 24 Jan · Shift 1 · Q53

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5D ? ['D' stands for dioptre]
  1. A
    0.04
  2. B
    0.1
  3. C
    0.40
  4. D
    0.01
View written solutionFree

Correct answer: A

  1. Use the relation between power and focal length

For a lens, P=1fP = \frac{1}{f}P=f1​ where PPP is in dioptres and fff is in metres.

  1. Initial power and new power

Initial power: P1=2.5 DP_1 = 2.5\,\text{D}P1​=2.5D

Increase in power: ΔP=0.1 D\Delta P = 0.1\,\text{D}ΔP=0.1D

So the new power is P2=2.5+0.1=2.6 DP_2 = 2.5 + 0.1 = 2.6\,\text{D}P2​=2.5+0.1=2.6D

  1. Find the corresponding focal lengths

Initial focal length: f1=12.5=0.4 mf_1 = \frac{1}{2.5} = 0.4\,\text{m}f1​=2.51​=0.4m

New focal length: f2=12.6≈0.3846 mf_2 = \frac{1}{2.6} \approx 0.3846\,\text{m}f2​=2.61​≈0.3846m

  1. Calculate the decrease in focal length

Δf=f1−f2=0.4−0.3846=0.0154 m\Delta f = f_1 - f_2 = 0.4 - 0.3846 = 0.0154\,\text{m}Δf=f1​−f2​=0.4−0.3846=0.0154m

  1. Relative decrease in focal length

Relative decrease means Δff1=0.01540.4=0.0385\frac{\Delta f}{f_1} = \frac{0.0154}{0.4} = 0.0385f1​Δf​=0.40.0154​=0.0385

Thus, relative decrease≈0.04\text{relative decrease} \approx 0.04relative decrease≈0.04

  1. Check options
  • A: 0.040.040.04 ✅
  • B: 0.10.10.1 ❌
  • C: 0.400.400.40 ❌
  • D: 0.010.010.01 ❌

Therefore, the correct option is A.

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