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  • Consider the situation shown in figure. Water  <img alt="\mathrm{\left(\mu_w=\frac{4}{3}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28%5Cmu_w%3D%5Cf

Consider the situation shown in figure. Water  \mathrm{\left(\mu_w=\frac{4}{3}\right)}  is filled in a breaker upto a height of 10 cm. A plane mirror is fixed at a height of 5 cm from the surface of water. Distance of the image of an object O at the bottom of the beaker formed by the mirror after reflection from it is

Option: 1

15 cm


Option: 2

12.5 cm


Option: 3

7.5 cm


Option: 4

10 cm


Answers (1)

best_answer

\mathrm{D_{\text {apparent }}=\frac{D_{\text {actual }}}{\mu_{\text {rel }}}=\frac{10}{\frac{4}{3}}=7.5 \mathrm{~cm}}

So, the object appears to be raised by 10 cm - 7.5 cm  =2.5 cm Hence, distance between mirror and  \mathrm{\mathrm{O}^{\prime}=5+7.5=12.5 \mathrm{~cm}}  So, final image will be formed at 12.5 cm behind the plane mirror.

Posted by

Sanket Gandhi

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