Two lenses of focal length $f_{1}$ and $f_{2}$ are kept in contact coaxially. The power of the combination will be

\frac{f_{1} f_{2}}{f_{1} + f_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{f_{1} + f_{2}}{f_{1} f_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{f_{1} f_{2}}{f_{1} - f_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f_{1} + f_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{f_{1} + f_{2}}{f_{1} f_{2}}$
If $f$ is focal length of combination
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
We know that $\frac{1}{f} = P$
So, $P = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$= \frac{f_{2} + f_{1}}{f_{1} f_{2}}$
$P = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$
Hence, option $(2)$ is correct.

Suggest Corrections

Similar questions

Two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are placed in contact with each other. The focal length of the combination will be given by:

If two thin lenses of focal length f₁ and f₂ are held in contact then the focal length of the combination will be

f_{1}

f_{2}

f_{1} a n d f_{2}

d < f_{1}, d < f_{2}

Δ

f_{1}

f_{2}

Join BYJU'S Learning Program