Question

A convex lens of focal length 50 cm and a concave lens of focal length 100 cm are separated by a distance of 10 cm what is the power of the combination

Answer 1

convex lens (f1) = -50 cm
concave lens (f2)= 100 cm

the equidistant focal length of combination is above

Answer 2

Concept:

Because the rays converge after passing through the convex lens, convex lenses are also known as converging lenses. Because the rays diverge after passing through the concave lens, it is called a diverging lens.

Find:

The power of the lens combination.

Solution:

Convex lens focal length $f_1=-50cm$.

Concave lens focal length $f_2=100cm$

Effective focal length

$\frac{1}{f} =\frac{1}{f_1} +\frac{1}{f_2}+\frac{a}{f_1f_2} \\\\\frac{1}{f} =-\frac{1}{50} +\frac{1}{100}-\frac{10}{5000} \\\\\frac{1}{f} =-0.012cm^{-1}$

Power

$P=-\frac{1}{f} \\P=0.012D$

Hence, the power of the combination is $0.012D$.

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