Question

Locate the image of the point P as seen by the eye in the figure.
Figure

Answer 1

I would be easy to answer if provide the pic of ur q ....

Answer 2

The point P as seen by the eye in the figure is 0.2 cm.    

Explanation:

Given data in the question

$\mu_{1}=1.2$

$\mu_{2}=1.3$

$\mu_{3}=1.4$

$t_{1}=0.2 \mathrm{cm}$

$t_{2}=0.3 \mathrm{cm}$

$t_{3}=0.4 \mathrm{cm}$

The presence of air in between the sheets has no effect on the change. The change is due to three sheets of different refractive index other than air.

  $\Delta t=\left(1-\frac{1}{\mu_{1}}\right) t_{1}+\left(1-\frac{1}{\mu_{2}}\right) t_{2}+\left(1-\frac{1}{\mu_{3}}\right) t_{3}$

        $=\left(1-\frac{1}{1.2}\right) \times 0.2+\left(1-\frac{1}{1.3}\right) \times 0.3+\left(1-\frac{1}{1.4}\right) \times 0.4$

        $=\left(\frac{1.2-1}{1.2}\right) \times 0.2+\left(\frac{1.3-1}{1.3}\right) \times 0.3+\left(\frac{1.4-1}{1.4}\right) \times 0.4$

        $=\left(\frac{0.2}{1.2}\right) \times 0.2+\left(\frac{0.3}{1.3}\right) \times 0.3+\left(\frac{0.4}{1.4}\right) \times 0.4$

        $=\left(\frac{1}{6}\right) \times 0.2+\left(\frac{3}{13}\right) \times 0.3+\left(\frac{4}{14}\right) \times 0.4$

        $=0.166 \times 0.2+0.230 \times 0.3+0.285 \times 0.4$

        = 0.033+0.069+0.114

        = 0.1022+0.114

        = 0.216

Therefore, the position of Image P is 0.2 cm above point P.