Optics Question 946
Question: A prism of dispersive power 0.021 and refractive index 1.53 form an achromatic combination with prism of angle $ 4.2{}^\circ $ and dispersive power 0.045 having refractive index 1.65. Find the resultant deviation.
Options:
A) $ 1.12{}^\circ $
B) $ 2.16{}^\circ $
C) $ 3.12{}^\circ $
D) $ 4.18{}^\circ $
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Answer:
Correct Answer: C
Solution:
[c] Here, $ \omega =0.021;\mu =1.53;\omega ‘=0.045; $ $ \mu ‘=1.65; $
$ A’={{4.2}^{o}} $
For no dispersion, $ \omega \delta + \delta’ \delta’ = 0 $
or $ \omega A(\mu -1)+\omega ‘A’(\mu -1)=0 $
or $ A=-\frac{\omega ‘A’(\mu ‘-1)}{\theta (\mu -1)} $
$ =\frac{0.045\times 4.2(1.65-1)}{0.02\times (1.53-1)}=-11.04$
Net deviation,
$ \delta +\delta ‘=A(\mu -1)+A’(\mu ‘-1) $
$ =-11.04(1.53-1)+4.2(1.65-1) $
$ =-11.04\times 0.53+4.2\times 0.65 $
$ =-5.85+2.73={{-3.12}^{o}} $
BETA
