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Question: Consider a concave mirror and a convex lens (refractive index \( = 1.5\)) of focal length \(10{\text...

Consider a concave mirror and a convex lens (refractive index =1.5 = 1.5) of focal length 10cm10{\text{cm}} each, separated by a distance of 50cm50{\text{cm}} in air (refractive index =1 = 1) as shown in the figure. An object is placed at a distance of 15cm15{\text{cm}} from the mirror. Its erect image formed by this combination has magnification M1{{\text{M}}_{\text{1}}}. When the set-up is kept in a medium of refractive index 76\dfrac{7}{6}, the magnification becomes M2{{\text{M}}_{\text{2}}}. The magnitude M2M1\left| {\dfrac{{{M_2}}}{{{M_1}}}} \right| is

Explanation

Solution

The first image from reflection by a spherical concave mirror can be found from mirror formula.

Formula used:
1v+1u=1f 1v1u=1f m=vu M=m1×m2  \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\\ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\\ m = \dfrac{v}{u} \\\ M = {m_1} \times {m_2} \\\

Complete step by step solution:
Given data,
For concave mirror
u=15cm, f=10cmu = - 15cm,{\text{ f}} = - 10cm
As we know,
1v+1u=1f\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}
1v+115=110 1v=110+115=1015150 v=30cm  \therefore \dfrac{1}{v} + \dfrac{1}{{ - 15}} = \dfrac{1}{{ - 10}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 1}}{{10}} + \dfrac{1}{{15}} = \dfrac{{10 - 15}}{{150}} \\\ \Rightarrow v = - 30cm \\\
Dividing (1) by (2)
fliquidfair=1232×671 =74  \dfrac{{{f_{liquid}}}}{{{f_{air}}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{3}{2} \times \dfrac{6}{7} - 1}} \\\ = \dfrac{7}{4} \\\
fliquid=74×10 =704cm  {f_{liquid}} = \dfrac{7}{4} \times 10 \\\ = \dfrac{{70}}{4}cm \\\
The first image will act as an object for lens and its position can be found using lens formula,
1v1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Given:
u=20cm, f=704cm   u = - 20cm, {\text{ f}} = \dfrac{{70}}{4}cm \\\ \\\
1v120=1704\therefore \dfrac{1}{v} - \dfrac{1}{{ - 20}} = \dfrac{1}{{\dfrac{{70}}{4}}}
1v×470=80701400\Rightarrow \dfrac{1}{v} \times \dfrac{4}{{70}} = \dfrac{{80 - 70}}{{1400}}
v=140cm\Rightarrow v = 140cm
Magnification is given by m=vu=14020=7m' = \dfrac{v}{u} = \dfrac{{140}}{{ - 20}} = - 7
Total magnification m2=2×7=14{m_2} = - 2 \times - 7 = 14

Hence, m2m1=142=7\left| {\dfrac{{{m_2}}}{{{m_1}}}} \right| = \dfrac{{14}}{2} = 7

Note: Focal length of lens depends on the medium in which the whole experiment is done. Students must be careful while taking sign convention.