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Geometrical Optics question

2023 · 24 Jan · Shift 1 · Q68
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Geometrical Optics question

2023 · 24 Jan · Shift 1 · Q68

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refraction index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance x=‾x=\underline{\hspace{2cm}}x=​ cm, from concave lens. JEE Main 2023 (Online) 24th January Morning Shift Physics - Geometrical Optics Question 82 English
Numerical answer
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Correct answer: 120

  1. Find focal lengths of the two lenses

For a thin lens in air, lens maker’s formula is

1f=(μ−1)(1R1−1R2)\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(μ−1)(R1​1​−R2​1​)

with μ=1.75\mu=1.75μ=1.75 and radius magnitude R=30 cmR=30\,\text{cm}R=30cm.

Because each lens is plano-type, one surface is plane and the other has radius 30 cm30\,\text{cm}30cm.

  • For the plano-convex lens:
1fc=(1.75−1)(130−0)=0.75⋅130=140\frac{1}{f_c}=(1.75-1)\left(\frac{1}{30}-0\right)=0.75\cdot \frac{1}{30}=\frac{1}{40}fc​1​=(1.75−1)(301​−0)=0.75⋅301​=401​

So,

fconvex=+40 cmf_{\text{convex}}=+40\,\text{cm}fconvex​=+40cm
  • For the plano-concave lens:
fconcave=−40 cmf_{\text{concave}}=-40\,\text{cm}fconcave​=−40cm
  1. Use the fact that the object is at infinity

The lens that first receives parallel rays will form an image at its focal point.

From the figure/context, the lenses are separated by 40 cm40\,\text{cm}40cm, equal to the focal length magnitude of each lens. This means the first lens forms its image exactly at the position of the second lens.

Let the plano-convex lens be the first lens. Then parallel rays from infinity form an image at distance

40 cm40\,\text{cm}40cm

to its right, i.e. exactly where the concave lens is placed.

So for the concave lens, the object is effectively at its optical centre location, hence the rays incident on it are converging toward a point at the lens plane itself.

  1. Image formation by the concave lens

Use the thin lens formula:

1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f}v1​−u1​=f1​

For the concave lens,

f=−40 cmf=-40\,\text{cm}f=−40cm

and the object is at the lens plane, so

u→0−u\to 0^{-}u→0−

More physically, just before reaching the concave lens, the rays would have converged at the lens plane; after refraction by a diverging lens of focal length −40-40−40 cm, they emerge as if coming from its focal point on the left.

Hence the final image is formed at

40 cm40\,\text{cm}40cm

from the concave lens on the same side as the incoming light.

However, depending on the figure orientation, if the concave lens is first and convex lens second, then:

  • concave lens receives parallel rays and forms a virtual image at 40 cm40\,\text{cm}40cm to its left,
  • this acts as object for the convex lens placed 40 cm40\,\text{cm}40cm away,
  • so for the convex lens the object distance is
u=−80 cmu=-80\,\text{cm}u=−80cm

(using Cartesian sign convention), and

1v−1(−80)=140\frac{1}{v}-\frac{1}{(-80)}=\frac{1}{40}v1​−(−80)1​=401​ 1v+180=140\frac{1}{v}+\frac{1}{80}=\frac{1}{40}v1​+801​=401​ 1v=140−180=180\frac{1}{v}=\frac{1}{40}-\frac{1}{80}=\frac{1}{80}v1​=401​−801​=801​ v=80 cmv=80\,\text{cm}v=80cm

Thus measured from the concave lens, the final image distance is

x=40+80=120 cmx=40+80=120\,\text{cm}x=40+80=120cm
  1. Final answer

Therefore,

x=120 cm\boxed{x=120\,\text{cm}}x=120cm​
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