JEE Advanced 2023 · Paper 1 · Physics Q9

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PhysicsInteger+4Question 26 of 51

An optical arrangement consists of two concave mirrors $M_1$ and $M_2$ , and a convex lens L with a common principal axis, as shown in the figure. The focal length of L is 10 cm. The radii of curvature of $M_1$ and $M_2$ are 20 cm and 24 cm, respectively. The distance between L and $M_2$ is 20 cm. A point object S is placed at the mid-point between L and $M_2$ on the axis. When the distance between L and $M_1$ is $n/7$ cm, one of the images coincides with S. The value of $n$ is _____.

Along a horizontal principal axis, from left to right: concave mirror M₁ (facing right), convex lens L, point object S, and concave mirror M₂ (facing left). L and M₂ are 20 cm apart, and S is midway between them.

Your answer

Answer and worked solution

Official answer: 80

Data: $f_L=10$ cm, $f_1=10$ cm ( $M_1$ ), $f_2=12$ cm ( $M_2$ ). S is 10 cm from both L and $M_2$ . Let $D$ be the distance from L to $M_1$ .

Light from S that first hits $M_2$ .

$M_2$ : $u=-10$ , $f=-12$ , so $\frac1v=-\frac1{12}+\frac1{10}=\frac1{60}$ , giving $v=+60$ .
This virtual image lies 60 cm behind $M_2$ , i.e. 80 cm from L.
L: object at 80 cm gives $\frac1v=\frac1{10}-\frac1{80}$ , so $v=\frac{80}{7}$ cm on the $M_1$ side.

Placing $M_1$ so the light returns to S.

Image at the pole of $M_1$ . $D=\frac{80}7$ . The light reflects back on itself through L and $M_2$ and re-forms at S: $n=80$ .
Image at the centre of curvature of $M_1$ . $D=\frac{80}7+20=\frac{220}7$ . The rays retrace their path back to S: $n=220$ .
Image at the focus of $M_1$ . $D=\frac{80}7+10=\frac{150}7$ . $M_1$ sends back a parallel beam, which L focuses at its focal point, 10 cm away, which is exactly S: $n=150$ .

The official key accepts all three values, 80, 150 or 220.

Topics:Combination of lenses and mirrors Spherical mirrors Thin lenses and lens maker's formula

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JEE Advanced 2023 · Paper 1, Physics question 9