The given lens is broken into four parts and rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is -

Question

The given lens is broken into four parts and rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is - (A) f  (B) f / 2 (C)  f / 4 (D) 4 f

Tardigrade Admin · Accepted Answer

Cutting a lens in transverse direction doubles their focal length i.e. 2 f. Using the formula of equivalent focal length, (1/f)=(1/f1)+(1/f2)+(1/f3)+(1/f4) We get equivalent focal length as f / 2.

Q. The given lens is broken into four parts and rearranged as shown. If the initial focal length is $f$ then after rearrangement the equivalent focal length is -

$f$

$f /2$

$f /4$

$4 f$

Cutting a lens in transverse direction doubles their focal length i.e. $2 f$ .
Using the formula of equivalent focal length,
$\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} + \frac{1}{f _{3}} + \frac{1}{f _{4}}$
We get equivalent focal length as $f /2$ .

Q. The given lens is broken into four parts and rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is -

f

f/2

f/4

4f

Cutting a lens in transverse direction doubles their focal length i.e. 2f. Using the formula of equivalent focal length, f1​=f1​1​+f2​1​+f3​1​+f4​1​ We get equivalent focal length as f/2.

Q. The given lens is broken into four parts and rearranged as shown. If the initial focal length is $f$ then after rearrangement the equivalent focal length is -

$f$

$f /2$

$f /4$

$4 f$

Cutting a lens in transverse direction doubles their focal length i.e. $2 f$ .
Using the formula of equivalent focal length,
$\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} + \frac{1}{f _{3}} + \frac{1}{f _{4}}$
We get equivalent focal length as $f /2$ .