A thin cylindrical rod of length 10 ~cm is placed horizontally on the principle axis of a concave mirror of focal length 20 ~cm . The rod is placed in a such a way that mid point of the rod is at 40 ~cm from the pole of mirror. The length of the image formed by the mirror will be \fracx3 ~cm . The value of x is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Main 2023 · 1 Feb · Shift 1 · Q66 with Solution

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Correct answer: 32

Understand the geometry

A cylindrical rod of length $10\,\text{cm}$ lies along the principal axis of a concave mirror.

Midpoint of rod is $40\,\text{cm}$ from the pole.
So its ends are at: $u_1 = 40-5 = 35\,\text{cm}, \qquad u_2 = 40+5 = 45\,\text{cm}$

Using Cartesian sign convention for mirrors:

concave mirror focal length: $f = -20\,\text{cm}$
object distances: $u_1=-35\,\text{cm},\; u_2=-45\,\text{cm}$

Use mirror formula

For each end of the rod, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

For the nearer end ( $u=-35$ cm):

$\frac{1}{-20} = \frac{1}{v_1} + \frac{1}{-35}$ $-\frac{1}{20} = \frac{1}{v_1} - \frac{1}{35}$ $\frac{1}{v_1} = -\frac{1}{20} + \frac{1}{35} = \frac{-7+4}{140} = -\frac{3}{140}$ $v_1 = -\frac{140}{3}\,\text{cm}$

For the farther end ( $u=-45$ cm):

$\frac{1}{-20} = \frac{1}{v_2} + \frac{1}{-45}$ $-\frac{1}{20} = \frac{1}{v_2} - \frac{1}{45}$ $\frac{1}{v_2} = -\frac{1}{20} + \frac{1}{45} = \frac{-9+4}{180} = -\frac{1}{36}$ $v_2 = -36\,\text{cm}$

Find image length

The image of the rod extends from $v_1$ to $v_2$ , so its length is $|v_1-v_2| = \left| -\frac{140}{3} - (-36) \right|$ $= \left| -\frac{140}{3} + \frac{108}{3} \right| = \left| -\frac{32}{3} \right| = \frac{32}{3}\,\text{cm}$

Given image length $= \dfrac{x}{3}\,\text{cm}$ , $\frac{x}{3} = \frac{32}{3}$ So, $x=32$

Comparison with stored answer

Derived answer is $32$ , which matches the stored correct answer.

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For the nearer end ( $u=-35$ cm):

For the farther end ( $u=-45$ cm):

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For the farther end ( $u=-45$ cm):

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For the farther end (u=−45u=-45u=−45 cm):

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For the farther end (u=−45u=-45u=−45 cm):

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For the nearer end ( $u=-35$ cm):

For the farther end ( $u=-45$ cm):