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  4. A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μ1 is put between the lens and the mirror, The pin has to be moved to A', such that OA' = 27 cm, to get its inverted real image at A' itself. The value of μ1 will be :

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Q. A thin convex lens $L$ (refractive index = 1.5) is placed on a plane mirror $M$. When a pin is placed at $A$, such that $OA = 18 \,cm$, its real inverted image is formed at $A$ itself, as shown in figure. When a liquid of refractive index $\mu_1$ is put between the lens and the mirror, The pin has to be moved to $A'$, such that $OA' = 27\, cm$, to get its inverted real image at $A'$ itself. The value of $\mu_1$ will be :

JEE MainJEE Main 2019Ray Optics and Optical Instruments

Solution:

$\frac{1}{f_{1}} = \frac{1}{2} \times\frac{2}{18} = \frac{1}{18} $
$ \frac{1}{f_{2}} = \frac{\left(\mu_{1} -1\right)}{-18} $
when $\mu_1$ is filled between lens and mirror
$ P= \frac{2}{18} - \frac{2}{18} \left(\mu_{1} - 1\right) = \frac{2-2\mu_{1} +2}{18} $
$ =F_{m} =- \left(\frac{18}{2-\mu_{1}}\right) $
$ 2=6-3\mu_{1} $
$ 3\mu_{1} = 4 $
$ \mu_{1} =4/3 $