A) \[3.3\text{ }cm\]
B) \[\text{2 }cm\]
C) \[\text{1}\text{.5 }cm\]
D) \[\text{4}\text{.2 }cm\]
Correct Answer: A
Solution :
[a] Image of the front face is formed at a distance of 40 cm on the other side of the lens. The rear face of the glass slab appears to be shifted by \[s=t\left( 1-\frac{1}{\mu } \right)=6\left( 1-\frac{1}{3/2} \right)=2cm\] nearer to the lens. For lens the distance of rear face of the slab is \[40+6-2=44\,cm.\]Its image is formed at a distance v from lens given by \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{v}-\frac{1}{-44}=\frac{1}{20}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{v}=\frac{1}{20}-\frac{1}{44}=\frac{11-5}{220}=\frac{3}{110}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,v=\frac{110}{3}=36.7\,cm\] \[\therefore \] Thickness of slab in the image formed by the lens is\[=40-36.7=3.3\text{ }cm\].You need to login to perform this action.
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