Jee Main 2024 01 02 2024 Shift 1 - Question23
Question 23
Distance between virtual magnified image, (size three times of object) of an object placed in front of convex lens and object is $20 cm$. The focal length of lens is $x cm$, then $x$ is
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Answer: (15)
Solution:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad \frac{v}{u}=3$
$3x - x = 20$
$x=20$
$\frac{1}{-30}-\frac{1}{-10}=\frac{1}{f}$
$\frac{1}{10}-\frac{1}{30}=\frac{1}{f}$
$\frac{2}{30}=\frac{1}{f} \Rightarrow f=15$
BETA
