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JEE Main 2021 (17 Mar, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2021 (17 Mar, Shift 2)

Q75JEE Main 2021 · 17 MAR · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at 23\frac{2}{3}rd of the distance of the object from the surface. The wavelength of light inside the surface is 23\frac{2}{3} times the wavelength in air. The radius of the curved surface is x13\frac{x}{13} m. The value of 'x' is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 30

  1. μ=λairλglass=32\mu = \frac{\lambda_{air}}{\lambda_{glass}} = \frac{3}{2}. Image v=+10v = +10 m, object at 32×10=15\frac{3}{2} \times 10 = 15 m, so u=−15u = -15 m.
  2. 1.510+115=0.5R\frac{1.5}{10} + \frac{1}{15} = \frac{0.5}{R}
  3. 9+460=1360=0.5R⇒R=3013\frac{9 + 4}{60} = \frac{13}{60} = \frac{0.5}{R} \Rightarrow R = \frac{30}{13} m
  4. So x=30x = 30.
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