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JEE Main 2022 (29 Jun, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2022 (29 Jun, Shift 1)

Q63JEE Main 2022 · 29 JUN · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is _____________ mm.

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

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Answer: 225

CFparallel beam22.5 cmR = 15 cm
  1. Sphere radius R=15R = 15 cm, μ=1.5\mu = 1.5. First surface (u=∞u = \infty): 1.5v1=0.515⇒v1=45\frac{1.5}{v_1} = \frac{0.5}{15} \Rightarrow v_1 = 45 cm.
  2. This is 45−30=1545 - 30 = 15 cm beyond the second surface, so for it u2=+15u_2 = +15 cm, R2=−15R_2 = -15 cm.
  3. 1v2−1.515=1−1.5−15⇒1v2=130+110⇒v2=7.5\frac{1}{v_2} - \frac{1.5}{15} = \frac{1 - 1.5}{-15} \Rightarrow \frac{1}{v_2} = \frac{1}{30} + \frac{1}{10} \Rightarrow v_2 = 7.5 cm
  4. Distance from centre =15+7.5=22.5= 15 + 7.5 = 22.5 cm (same as μR2(μ−1)\frac{\mu R}{2(\mu - 1)}).
  5. So the answer is 225 mm.
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