A) \[1.12{}^\circ \]
B) \[2.16{}^\circ \]
C) \[3.12{}^\circ \]
D) \[4.18{}^\circ \]
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 \[A=-\frac{\omega 'A'(\mu '-1)}{\theta (\mu -1)}\] |
| \[=\frac{0.045\times 4.2(1.65-1)}{0.02\times (1.53-1)}=-{{11.04}^{o}}\] |
| Net deviation, |
| \[\delta +\delta '=A(\mu -1)+A'(\mu '-1)\] |
| \[=-11.04(1.53-1)+4.2(1.65-1)\] |
| \[=-11.04\times 0.53+4.2\times 0.65\] |
| \[=-5.85+2.73={{3.12}^{o}}\] |
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