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Geometrical Optics question

2020 · 8 Jan · Shift 1 · Q50
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Geometrical Optics question

2020 · 8 Jan · Shift 1 · Q50

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 60

  1. Given data

    • Plano-convex lens in air
    • Radius of curvature of curved surface: R=30 cmR = 30\,\text{cm}R=30cm
    • Refractive index of lens material: μ=1.5\mu = 1.5μ=1.5

  2. Use 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​)

  3. For a plano-convex lens

    • One surface is plane, so its radius is infinite: R2=∞R_2 = \inftyR2​=∞
    • The other surface is curved with radius R1=30 cmR_1 = 30\,\text{cm}R1​=30cm

    Therefore,

    1f=(1.5−1)(130−1∞)\frac{1}{f} = (1.5 - 1)\left(\frac{1}{30} - \frac{1}{\infty}\right)f1​=(1.5−1)(301​−∞1​)

  4. Simplify

    1f=0.5⋅130=160\frac{1}{f} = 0.5 \cdot \frac{1}{30} = \frac{1}{60}f1​=0.5⋅301​=601​

    Hence,

    f=60 cmf = 60\,\text{cm}f=60cm

  5. Final answer The focal length of the lens is

    60\boxed{60}60​

  6. Comparison with stored answer

    • Derived answer: 606060
    • Stored correct answer: 606060

    They match.

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