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Q.
The image of the point $P$ as seen by the eye in the figure above $P$ at $0.11 \times x\, cm$ then the value of $x$ is ?
Ray Optics and Optical Instruments
Solution:
Apparent depth of $P$ as seen by observer
$=\left(\frac{t_{1}}{\mu_{1}}+\frac{t_{2}}{\mu_{2}}+\frac{t_{3}}{\mu_{3}}+\frac{t_{4}}{\mu_{4}}+\frac{t_{5}}{\mu_{5}}+\frac{t_{6}}{\mu_{6}}\right) \mu_{7} $
$=\left(\frac{1}{1}+\frac{0.2}{1.2}+\frac{1}{1}+\frac{0.3}{1.3}+\frac{1}{1}+\frac{0.4}{1.4}\right)(1)$
$=\left(3+\frac{1}{6}+\frac{3}{13}+\frac{2}{7}\right)=3.68\, cm$
So image of $P$ is formed at a distance $3.68\,cm$ below the top of slab of thickness $0.4\,cm$ or at a distance $(3.9-3.68)=0.22\, cm$ above $P$.
