A) \[f/2\]
B) \[3f/2\]
C) \[2f\]
D) \[f\]
Correct Answer: D
Solution :
\[u=-\left( f+\frac{f}{2} \right)=-\frac{3f}{2}\] Distance of final image from slab\[=\frac{3f}{2}\] Distance of image formed due to convex lens from face of the slab facing the lens. \[=\frac{(3/2)f}{\mu }=f\] Let \[{{F}_{2}}S=x\] \[v=2f+x\] From \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\frac{1}{(x+2f)}+\frac{1}{(-3f/2)}=\frac{1}{f}\]or\[x=f\]You need to login to perform this action.
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