A thin cylindrical rod of length 10 cm is placed horizontally on the principal axis of a concave mirror of focal length 20 cm. [2023]

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Solution

Q.
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 a such a way that the midpoint of the rod is at 40 cm from the pole of the mirror. The length of the image formed by the mirror will be $\frac{x}{3} m$ . The value of $x$ is _______. [2023]

Ans.
(32)

$U_{A} = - 45 cm, f = - 20 cm$

$V_{A} = \frac{- 45 \times (- 20)}{- 45 - (- 20)} = \frac{- 900}{25} = - 36 cm$

$And U_{B} = - 35 cm$

$∴$ $V_{B} = \frac{- 35 \times (- 20)}{- 35 - (- 20)} = \frac{700}{- 15} = - \frac{140}{3} cm$

$∴$ $V_{A} - V_{B} = length of image$

$= (- 36 + \frac{140}{3}) cm = \frac{- 108 + 140}{3} cm = \frac{32}{3} cm$

$∴$ $x = 32$

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