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MHT CET 2026 (15 May, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2026 (15 May, Shift 2)

Q8MHT CET 2026 · 15 MAY · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

A biconvex lens has radius of curvature of both the surfaces same as 16\frac{1}{6} cm. If this lens is replaced by another convex lens having different radii of curvature on both sides without change in lens power then possible combination of R1R_1 and R2R_2 is (R1≠R2R_1 \neq R_2)

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Answer: (b)

  1. Power ∝1R1+1R2\propto \frac{1}{R_1} + \frac{1}{R_2} for a biconvex lens of the same material.
  2. Original: 6+6=12 cm−16 + 6 = 12\ \text{cm}^{-1}.
  3. Check options: (a) 3+5=83 + 5 = 8, (b) 5+7=125 + 7 = 12, (c) 3+7=103 + 7 = 10, (d) 6+9=156 + 9 = 15.
  4. So the answer is (b) 15\frac{1}{5} cm and 17\frac{1}{7} cm.
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