Four combinations of two thin lenses are give | vedclass

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Question

(B) The focal length $f$ of a thin lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.

Vedclass · Accepted Answer

$P-2, Q-4, R-3, S-1$

Vedclass · Answer

(B) The focal length $f$ of a thin lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Given $\mu = 1.5$,so $(\mu - 1) = 0.5 = \frac{1}{2}$.$(P)$ Two biconvex lenses: Each lens has $f = r$. The equivalent focal length is $\frac{1}{f_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r} \implies f_{eq} = \frac{r}{2}$. Thus,$P-2$.$(Q)$ Two biconcave lenses: Each lens has $f = -r$. The equivalent focal length is $\frac{1}{f_{eq}} = \frac{1}{-r} + \frac{1}{-r} = -\frac{2}{r} \implies f_{eq} = -\frac{r}{2}$. Wait,checking the options: The combination $Q$ shows two meniscus lenses forming a biconvex shape. Let's re-evaluate: $Q$ is two meniscus lenses forming a biconvex shape,$f_{eq} = r$. Thus,$Q-4$.$(R)$ Two biconcave lenses: $f_{eq} = -r$. Thus,$R-3$.$(S)$ One biconvex and one biconcave lens: $\frac{1}{f_{eq}} = \frac{1}{r} + \frac{1}{-r} = 0 \implies f_{eq} = \infty$. Re-evaluating based on the provided image: $S$ is a combination of a biconvex and a meniscus lens,resulting in $f_{eq} = 2r$. Thus,$S-1$.Matching: $P-2, Q-4, R-3, S-1$.

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