Distance between virtual image, which is twice the size of object placed in front of mirror and object is 45 cm. The magnitude of focal length of the mirror is _____cm.
The Correct Answer is: 30 \(|m|=|\frac{v}{u}|=2\) \(|v|=|2u|\) Let the unknown quantities be n, and 2n. Their sum is 45. n + 2n = 45 Solving for n: 3n = 45 n = 15 cm Therefore, u = –15 cm and v = 30 cm. \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) Substitute the values of v and u: \(\frac{1}{30}+\frac{1}{-15}=\frac{1}{f}\) Calculate the reciprocal of the focal length f: \(\frac{1-2}{30}=\frac{-1}{30}=\frac{1}{f}\) Thus, the focal length is: \(⇒ f=30\ cm\)
