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JEE Main 2026 (2 Apr, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2026 (2 Apr, Shift 1)

Q16JEE Main 2026 · 2 APR · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

Refer the figure given below. μ1\mu_1 and μ2\mu_2 are refractive indices of air and lens material. The height of image will be _______ cm.

Figure

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Answer: (a)

  1. Refraction at a spherical surface: μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}, with u=−40u = -40 cm, R=−20R = -20 cm (C on the object side).
  2. 1.54v+140=0.54−20⇒1.54v=−0.052⇒v≈−29.6\frac{1.54}{v} + \frac{1}{40} = \frac{0.54}{-20} \Rightarrow \frac{1.54}{v} = -0.052 \Rightarrow v \approx -29.6 cm
  3. m=μ1vμ2u=−29.61.54×(−40)≈0.48m = \frac{\mu_1 v}{\mu_2 u} = \frac{-29.6}{1.54 \times (-40)} \approx 0.48
  4. Image height =0.48×2≈0.96≈1= 0.48 \times 2 \approx 0.96 \approx 1 cm. So the answer is (a) 1.
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