Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

Two thin lenses of focal lengths f1 and f2 are in contact and coaxial. The combination is equivalent to a single lens of power

Options:

$f_1 + f_2$

$\frac{f_1f_2}{f_1+f_2}$

$\frac{f_1+f_2}{2}$

$\frac{f_1+f_2}{f_1f_2}$

Correct Answer:

$\frac{f_1+f_2}{f_1f_2}$

Explanation:

$\text{Equivalent focal length is given by } \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$

$\Rightarrow F = \frac{f_1f_2}{f_1+f_2}$

Equivalent Power $ P = \frac{1}{F} = \frac{f_1+ f_2}{f_1f_2}$