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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2015 · Paper 1 · Physics · Ray Optics

Consider a concave mirror and a convex lens (refractive index = 1 ....

Consider a concave mirror and a convex lens (refractive index = 1.51 . 5) of focal length 10cm10 cm each, separated by a distance of 50cm50 cm in air (refractive index = 11), as shown in the figure. An object is placed at a distance of 15cm15 cm from the mirror. Its erect image, formed by this combination, has magnification M1{M}_{1} . When the set-up is kept in a medium of refractive index 76\dfrac{7}{6} , the magnification becomes M2{M}_{2}. The magnitude \left|\dfrac{{M}_{2}}{{M}_{1}}\right|$ is
Consider a concave mirror and a convex lens (refractive index = 1 . 5 ) of focal length 10 cm each, separated by a

Correct answer

7

Step-by-step solution

For reflection from a concave mirror,
1v+1u=1f⇒1v−115=−110\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} ⇒ \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{- 1}{10}
1v=115−110=−130\dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{10} = \dfrac{- 1}{30}
∴v=−30∴ v = - 30
Magnification \left({m}_{1}\right) = - \dfrac{v}{u} = - 2<br/>Nowforrefractionfromlens,<br/><br />Now for refraction from lens,<br />\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} ⇒ \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{20} = \dfrac{1}{20}<br/><br />∴Magnification(m2)=vu=−1Magnification \left({m}_{2}\right) = \dfrac{v}{u} = - 1
∴M1=m1m2=2∴ {M}_{1} = {m}_{1}{m}_{2} = 2
Now when the set-up is immersed in liquid, no effect for the image formed by mirror.
We have (μL−1)(1R1−1R2)=110\left({\mu}_{L} - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) = \dfrac{1}{10}
⇒(1R1−1R2)=15⇒ \left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) = \dfrac{1}{5}
When lens is immersed in liquid,
1flens=(μLμS−1)(1R1−1R2)=27×15=235\dfrac{1}{{f}_{lens}} = \left(\dfrac{{\mu}_{L}}{{\mu}_{S}} - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) = \dfrac{2}{7} \times \dfrac{1}{5} = \dfrac{2}{35}
∴1v−1u=1fLiquid∴ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{{f}_{Liquid}}
⇒1v=235−120=8−7140=1140⇒ \dfrac{1}{v} = \dfrac{2}{35} - \dfrac{1}{20} = \dfrac{8 - 7}{140} = \dfrac{1}{140}
∴∴ Magnification =−14020=−7= - \dfrac{140}{20} = - 7
∴M2=2×7=14∴ {M}_{2} = 2 \times 7 = 14
∴∣M2M1∣=7∴ \left|\dfrac{{M}_{2}}{{M}_{1}}\right| = 7
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