Optics Ques 75
- Four combinations of two thin lenses are given in Column I.The radius of curvature of all curved surfaces is $r$ and the refractive index of all the lenses is $1.5$. Match lens combinations in Column I with their focal length in Column II and select the correct answer using the codes given below the lists.
(2014 Adv.)
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Answer:
Correct Answer: 75.(A) $\rightarrow$ p; (B) $\rightarrow s$;
(C) $\rightarrow r$; (D) $\rightarrow p$
Solution:
Formula:
- (P) $ \frac{1}{f}=(\frac{3}{2}-1) (\frac{1}{r}+\frac{1}{r})=\frac{1}{r} \Rightarrow f=r$
$ W \Rightarrow \quad \frac{1}{f _{eq}}=\frac{1}{f}+\frac{1}{f}=\frac{2}{r} \Rightarrow f _{eq}=\frac{r}{2} $
(Q) $ \quad \frac{1}{f}=(\frac{3}{2}-1) (\frac{1}{r}) \Rightarrow f=2 r$
$M \Rightarrow \frac{1}{\phi}+\frac{1}{\phi}=\frac{2}{\phi}=\frac{1}{\rho} \Rightarrow f _{\text {eq }}=r$
(R) $ \frac{1}{f}=(\frac{3}{2}-1) \quad(-\frac{1}{r})=-\frac{1}{2 r} \Rightarrow f=-2 r$
$ \Rightarrow \frac{1}{f _{eq}}=\frac{1}{f}+\frac{1}{f}=-\frac{2}{2 r} \Rightarrow f _{eq}=-r $
(S) $\Rightarrow \frac{1}{f _{\text {eq }}}=\frac{1}{r}+\frac{1}{-2 r}=\frac{1}{2 r} $ $\Rightarrow f _{\text {eq }}=2 r.$
BETA
