A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification m₁ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed, a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to m₂. The value of $\left|\frac{{m}_{1}}{{m}_{2}}\right|$ is ____.

Question

YESPYQ · Accepted Answer

Correct answer: a) $\frac{5}{6}$. In Case-1: image is formed at $\frac{20}{3}$ cm from combination $\Rightarrow \left|{m}_{1}\right|=\frac{20/3}{10}=\frac{2}{3}$ In Case-2: Image formed at $\frac{36}{5}$ cm from concave lens $\Rightarrow \left|{m}_{\text{concave}}\right|=\frac{36/5}{9}=\frac{4}{5}$ $\left|{m}_{\text{convex}}\right|=1$ $\Rightarrow \left|{m}_{2}\right|=\frac{4}{5}$ $\frac{\left|{m}_{1}\right|}{\left|{m}_{2}\right|}=\frac{5}{6}$

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any …

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