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

Question

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 x 3 cm . The value of x is ______.

TestHub · Accepted Answer

As per sign conventions, object distance $u_{A} = - 45 cm$ , focal length $f = - 20 cm$ .

From the mirror formula

$\frac{1}{v_{A}} + \frac{1}{u_{A}} = \frac{1}{f} \Rightarrow \frac{1}{v_{A}} + \frac{1}{- 45} = \frac{1}{- 20} \Rightarrow v_{A} = \frac{- 45 \times (- 20)}{- 45 - (- 20)} = \frac{- 900}{25} \Rightarrow v_{A} = - 36 cm$

And $u_{B} = - 35 cm$

$\frac{1}{v_{B}} + \frac{1}{u_{B}} = \frac{1}{f} \Rightarrow \frac{1}{v_{B}} + \frac{1}{- 35} = \frac{1}{- 20} \Rightarrow v_{B} = \frac{- 35 \times (- 20)}{- 35 - (- 20)} = \frac{700}{- 15} \Rightarrow v_{B} = - \frac{140}{3} cm$

Now, length of image $= v_{A} - v_{B}$

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

$= \frac{- 108 + 140}{3} cm$

$= \frac{32}{3} cm$

$∴ x = 32$

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