Question

An object is kept at 10 cm in front of a convex
lens. Its image is formed on the screen at
15 cm from the lens. Calculate (1) the focal
length of the lens. (2) the magnification
produced by the lens.​

Answer 1

Answer:

We know that,

$\frac{1}{f}=\frac{1}{v} - \frac{1}{u}$

$\frac{1}{f} = \frac{1}{15} - \frac{1}{-10}$ ( As u is always negative)

$\frac{1}{15} + \frac{1}{10} \\\frac{2+3}{30} = \frac{5}{30} = \frac{1}{6} \\\frac{1}{f} = \frac{1}{6} \\f=6$

The magnification in lens = $\frac{v}{u} = \frac{+15}{-10} = -1.5$

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