A) 4.56 cm
B) 5.65 cm
C) 6.56 cm
D) 7.67 cm
Correct Answer: D
Solution :
\[u=21cm;\] \[f=\frac{R}{2}=\frac{10}{2}=5\,\,cm\] |
On introducing the glass slab, the object as well as the image will be shifted from the mirror through a distance |
\[d=t\left( 1-\frac{1}{\mu } \right)=3\left( 1-\frac{1}{1.5} \right)=1\,cm,\] |
So, that apparent distance of the object =20 cm. (i.e.) u = 20 cm. |
By the mirror formula \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] |
\[v=\frac{20}{3}cm=6.67\,\,cm\] |
Distance of the final image from the mirror \[=6.67+1=7.67\,\,cm.\] |
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