Ray Optics Question 13

Question 13 - 01 February - Shift 1

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 $\frac{x}{3} cm$. The value of $x$ is

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Answer: (32)

Solution:

Formula: Mirror Formula

$U_A=-45 cm, f=-20 cm$

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

And $U_B=-35 cm$

$\therefore V_B=\frac{-35 \times(-20)}{-35-(-20)}=\frac{700}{-15}$

$\therefore V_A-V_B=$ length of image

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

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

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

$\therefore x=32$