A) 12°2.4'
B) 12°4?
C) 1.24°
D) 12°
Correct Answer: A
Solution :
For dispersion without deviation \[\frac{A}{{{A}'}}=\left( \frac{{{{{\mu }'}}_{y}}-1}{{{\mu }_{y}}-1} \right)\] \[\therefore \]\[\frac{A}{10}=\frac{(1.602-1)}{(1.500-1)}=\frac{0.602}{0.500}\] Þ \[A=12{}^\circ 2.4'\]You need to login to perform this action.
You will be redirected in
3 sec