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Geometrical Optics question

2025 · 3 Apr · Shift 1 · Q67
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Geometrical Optics question

2025 · 3 Apr · Shift 1 · Q67

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm . The refractive index of the lens material is
  1. A
    1.4
  2. B
    1.8
  3. C
    1.5
  4. D
    1.2
View written solutionFree

Correct answer: C

  1. Use the lens maker 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 biconvex lens, using the Cartesian sign convention:

  • First surface: R1=+10 cmR_1=+10\,\text{cm}R1​=+10cm
  • Second surface: R2=−15 cmR_2=-15\,\text{cm}R2​=−15cm
  • Focal length: f=+12 cmf=+12\,\text{cm}f=+12cm
  1. Substitute the values:
112=(μ−1)(110−1−15)\frac{1}{12}=(\mu-1)\left(\frac{1}{10}-\frac{1}{-15}\right)121​=(μ−1)(101​−−151​) 112=(μ−1)(110+115)\frac{1}{12}=(\mu-1)\left(\frac{1}{10}+\frac{1}{15}\right)121​=(μ−1)(101​+151​)
  1. Add the fractions inside the bracket:
110+115=3+230=530=16\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}101​+151​=303+2​=305​=61​

So,

112=(μ−1)⋅16\frac{1}{12}=(\mu-1)\cdot \frac{1}{6}121​=(μ−1)⋅61​
  1. Solve for μ\muμ:
μ−1=112×6=12\mu-1=\frac{1}{12}\times 6=\frac{1}{2}μ−1=121​×6=21​ μ=1+12=1.5\mu=1+\frac{1}{2}=1.5μ=1+21​=1.5
  1. Match with the options:

1.51.51.5 corresponds to Option C.

Comparison with stored correct answer

Stored correct answer: C
Derived answer: C

So, the derived answer agrees with the stored correct answer.

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