Question

Two lens of power -15 day and +5d are are placed coaxially the focal length of equivalent lens is

Answer 1

The equivalent focal length of the lens is -10d

Answer 2

Given,

Power of two lenses are $P_{1}=-15 D$ and  $P_{2}=+5 D$

When two lenses are kept in contact, powers of the lenses are simply added.

Therefore the power of the equivalent lens $P_{3}=-15+5=-10 D$

We know that the the lens power is the inverse of the focal length in meters.

Consider the focal length of the equivalent lens = f meter.

Therefore, $f_{e}=\frac{1}{D_{3}}=\frac{1}{-10}=-0.1\text{ m}=-10\text{ cm}$

The equivalent lens is a concave lens, so the object and image are on the same side of the lens.

so, we can say that the focal length of equivalent lens is =10 cm