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QJEEAdv21Ph63f8: JEE Advanced 2021 Physics - Class 12 JEEAdv21Ph63f8 Solution: An extended object is placed at point O, 10 cm in front of a…

An extended object is placed at point O, 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses ae 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is

JEE Advanced 2021 Paper 1 Online Physics - Geometrical Optics Question 35 English

Multiple Choice Options

  • A 0.4
  • B 0.8
  • C 1.3
  • D 1.6

Step-by-Step Verified Solution

Focal length of convex lens f1,

Therefore 1f1=(μ−1)[1R1−1R2]=(1.5−1)[120−(1−20)]{1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right] = (1.5 - 1)\left[ {{1 \over {20}} - \left( {{1 \over { - 20}}} \right)} \right]

⇒\Rightarrow f1 = + 20 cm

Focal length of concave lens f2,

∴\therefore 1f2=(μ−1)[1R1−1R2]{1 \over {{f_2}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]

1f2=(1.5−1)[−120−120]=1−20{1 \over {{f_2}}} = (1.5 - 1)\left[ { - {1 \over {20}} - {1 \over {20}}} \right] = {1 \over { - 20}}

⇒\Rightarrow f2 = −- 20 cm

For lens 1

JEE Advanced 2021 Paper 1 Online Physics - Geometrical Optics Question 35 English Explanation
1v−1u=1f{1 \over v} - {1 \over u} = {1 \over f}

⇒\Rightarrow v = −- 20 cm

m1=vu=−20−10=2{m_1} = {v \over u} = {{ - 20} \over { - 10}} = 2

For lens 2

u = −- 30n cm, f = −- 20 cm,

1v−1u=1f{1 \over v} - {1 \over u} = {1 \over f}

v = −- 12 cm

m2=vu=−12−30=25{m_2} = {v \over u} = {{ - 12} \over { - 30}} = {2 \over 5}

Net magnification,

m=m1m2=2×25=45=0.8m = {m_1}{m_2} = 2 \times {2 \over 5} = {4 \over 5} = 0.8

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