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Geometrical Optics question

2023 · 13 Apr · Shift 2 · Q70
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Geometrical Optics question

2023 · 13 Apr · Shift 2 · Q70

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
A bi convex lens of focal length 10 cm10 \mathrm{~cm}10 cm is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ‾\underline{\hspace{2cm}}​ D.
Numerical answer
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Correct answer: 5

  1. Given: A bi-convex lens has focal length f=10 cm=0.10 mf = 10\text{ cm} = 0.10\text{ m}f=10 cm=0.10 m

  2. Original power of the lens: P=1f(in m)=10.10=10 DP = \frac{1}{f(\text{in m})} = \frac{1}{0.10} = 10\,\text{D}P=f(in m)1​=0.101​=10D

  3. Nature of the cut:
    The lens is cut into two identical parts by a plane perpendicular to the principal axis.
    This means the lens is split into two thinner plano-convex lenses, each having:

    • one curved surface same as original,
    • one plane surface.

  4. Using lens maker's formula: For the original bi-convex lens, P=(μ−1)(1R1−1R2)P = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)P=(μ−1)(R1​1​−R2​1​) For a symmetric bi-convex lens, R1=R,R2=−RR_1 = R, \quad R_2 = -RR1​=R,R2​=−R so

    = (\mu-1)\left(\frac{2}{R}\right)$$ After cutting perpendicular to the principal axis, each part becomes a plano-convex lens: $$P' = (\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) = (\mu-1)\frac{1}{R}$$ Hence, $$P' = \frac{P}{2}$$

  5. Compute new power: P′=102=5 DP' = \frac{10}{2} = 5\,\text{D}P′=210​=5D

  6. Final answer: Each cut lens has power 5 D\boxed{5\,\text{D}}5D​

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