The image of point P when viewed from top of the slabs will be :

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The image of point $P$ when viewed from top of the slabs will be :

Question :

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

Solution :

First method :

Using the formula of apparent depth for multiple mediums, i.e.,

$\displaystyle \frac{{{d}_{app}}_{.}}{{{\mu }_{obesever}}}=\frac{{{x}_{_{obesever}}}}{{{\mu }_{obesever}}}+\frac{{{x}_{1}}}{{{\mu }_{1}}}+\frac{{{x}_{2}}}{{{\mu }_{2}}}+...$

Here the observer is very close to the first interface, hence neglecting the term $\displaystyle \frac{{{x}_{_{obesever}}}}{{{\mu }_{obesever}}}$ , we get

$\displaystyle \frac{{{d}_{app}}_{.}}{{{\mu }_{obesever}}}=\frac{{{x}_{1}}}{{{\mu }_{1}}}+\frac{{{x}_{2}}}{{{\mu }_{2}}}+...$

$\displaystyle \Rightarrow \frac{{{d}_{app}}_{.}}{1}=\frac{1.5}{1.5}+\frac{1.5}{1}+\frac{1.5}{1.5}+\frac{2}{1}$

$\displaystyle \Rightarrow {{d}_{app}}_{.}=5.5\text{ }cm$

(here refractive index of glass is taken as 1.5)

Real depth of P = 6.5 cm

Apparent depth of P = 5.5 cm

Shift = 1 cm above P

Hence correct option is (d).

Second method :

Shift observed due to refraction through glass slab is given by :

$\displaystyle Shift\text{ }=\text{ }t\left( 1-\frac{1}{\mu } \right)$

As there are two slabs between the observer and object(Point P), hence total shift is given by :

$\displaystyle Shift\text{ }=\text{ 2}\times t\left( 1-\frac{1}{\mu } \right)=2\times 1.5\left( 1-\frac{1}{1.5} \right)=1\text{ }cm$

Shift = 1 cm above P

Hence correct option is (d).

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