Question

if one surface of a double convex lens has been silvered then its effective focal length f is​

Answer 1

Answer:

It's focal length becomes R/2u times

Explanation:

where R is radius and u is refractive index of lens

Answer 2

Effective focal length $F=\frac{f_{m} \times f_{t} }{2f_{m} \ + f_{t} }$

Given:

One surface of a double convex lens has been silvered.

To find:

Effective focal length f.

Explanation:

When silvering the surface of the lens it behaves like concave mirror.

The focal length of the silvered lens $\[\frac{1}{F}=\frac{2}{{{f}_{l}}}+\frac{1}{{{f}_{m}}}\]$

           Where.  $\[{{f}_{l}}=\]$ Focal length of lens from which refraction takes place

                        $\[{{f}_{m}}=\]$ Focal length of mirror from which reflection takes place.

       $\[\frac{1}{F}=\frac{2}{{{f}_{l}}}+\frac{1}{{{f}_{m}}}\]$

        $\[\frac{1}{F}=\frac{2f_{m} \ + f_{t} }{f_{m} \times f_{t} }$

        $F=\frac{f_{m} \times f_{t} }{2f_{m} \ + f_{t} }$

Effective focal length    $F=\frac{f_{m} \times f_{t} }{2f_{m} \ + f_{t} }$

To learn more...

1)Name the type of lence if its power is

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2)What is concave lence

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