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Geometrical Optics question

2024 · 6 Apr · Shift 2 · Q71
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Geometrical Optics question

2024 · 6 Apr · Shift 2 · Q71

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
For the thin convex lens, the radii of curvature are at 15 cm15 \mathrm{~cm}15 cm and 30 cm30 \mathrm{~cm}30 cm respectively. The focal length the lens is 20 cm20 \mathrm{~cm}20 cm. The refractive index of the material is :
  1. A
    1.2
  2. B
    1.5
  3. C
    1.4
  4. D
    1.8
View written solutionFree

Correct answer: B

  1. Use the lens maker’s formula for a thin lens in air:
1f=(μ−1)(1R1−1R2)\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(μ−1)(R1​1​−R2​1​)

For a convex lens, taking light from left to right, we use:

  • R1=+15 cmR_1 = +15\,\text{cm}R1​=+15cm
  • R2=−30 cmR_2 = -30\,\text{cm}R2​=−30cm
  • f=+20 cmf = +20\,\text{cm}f=+20cm
  1. Substitute the values:
120=(μ−1)(115−1−30)\frac{1}{20}=(\mu-1)\left(\frac{1}{15}-\frac{1}{-30}\right)201​=(μ−1)(151​−−301​) 120=(μ−1)(115+130)\frac{1}{20}=(\mu-1)\left(\frac{1}{15}+\frac{1}{30}\right)201​=(μ−1)(151​+301​) 120=(μ−1)(2+130)\frac{1}{20}=(\mu-1)\left(\frac{2+1}{30}\right)201​=(μ−1)(302+1​) 120=(μ−1)(330)\frac{1}{20}=(\mu-1)\left(\frac{3}{30}\right)201​=(μ−1)(303​) 120=(μ−1)(110)\frac{1}{20}=(\mu-1)\left(\frac{1}{10}\right)201​=(μ−1)(101​)
  1. Solve for μ\muμ:
μ−1=1/201/10=1020=12\mu-1=\frac{1/20}{1/10}=\frac{10}{20}=\frac{1}{2}μ−1=1/101/20​=2010​=21​ μ=1+12=1.5\mu=1+\frac{1}{2}=1.5μ=1+21​=1.5
  1. Check options:
  • A: 1.21.21.2 ❌
  • B: 1.51.51.5 ✅
  • C: 1.41.41.4 ❌
  • D: 1.81.81.8 ❌

Therefore, the refractive index of the lens material is 1.51.51.5.

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