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