Question

In the given figure the image of point P when viewed from top of the slabs will be :

• A. $2.0$cm above P
• B. $1.5$cm above P
• C. $2.0$cm below P
• D. $1$cm above P

Answer 1

Answer: (D)

$1$cm above P

The two slabs will shift the image a distance

$d=2(1-\frac{1}{\mu })t⟹d=2(1-\frac{1}{1.5})\left(1.5\right)⟹d=2\times \frac{0.5}{1.5}\times 1.5⟹d=1cm$

Therefore, final image will be 1 cm above point P.