A thin cylindrical rod of length 10 cm is placed horizontally on the principal axis of a concave mirror of focal length 20 cm. The rod is placed in 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 x/3 cm. The value of x is _____________. Numerical answer type. Work it out, then open the answer.

Answer: 32. f = -20 cm; the rod ends are at u = -35 cm and u = -45 cm. u = -35: 1/v = -1/20 + 1/35 = -3/140 ⇒ v = -140/3 cm u = -45: 1/v = -1/20 + 1/45 = -1/36 ⇒ v = -36 cm Image length = 140/3 - 36 = 32/3 cm ⇒ x = 32

A thin cylindrical rod of length 10 cm is placed horizontally on the… | JEE Main 2023 (1 Feb, Shift 1)

Q26JEE Main 2023 · 1 FEB · SHIFT 1Medium

Reflection & Mirrors

A thin cylindrical rod of length $10\text{ cm}$ is placed horizontally on the principal axis of a concave mirror of focal length $20\text{ cm}$ . The rod is placed in such a way that mid point of the rod is at $40\text{ cm}$ from the pole of mirror. The length of the image formed by the mirror will be $\frac{x}{3}\text{ cm}$ . The value of $x$ is _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 32

$f = -20\text{ cm}$ ; the rod ends are at $u = -35\text{ cm}$ and $u = -45\text{ cm}$ .
$u = -35$ : $\frac{1}{v} = -\frac{1}{20} + \frac{1}{35} = -\frac{3}{140} \Rightarrow v = -\frac{140}{3}\text{ cm}$
$u = -45$ : $\frac{1}{v} = -\frac{1}{20} + \frac{1}{45} = -\frac{1}{36} \Rightarrow v = -36\text{ cm}$
Image length $= \frac{140}{3} - 36 = \frac{32}{3}\text{ cm} \Rightarrow x = 32$

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