Physics - Geometrical Optics Question with Solution | TestHub

If a lens of focal length$f$is divided into two equal parts  and each has a focal length$f\prime$then

$\frac{1}{f}=\frac{1}{f\prime }+\frac{1}{f\prime }$, i.e,$f\prime =2f$

Now if these parts are put in contact, then resultant focal length of the combination will be

$\frac{1}{F}=\frac{1}{2f}+\frac{1}{2f}$,i.e,$F=f$(initial value)

For this combination,

$\frac{1}{F}=\left({ }_a{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \dots \dots \dots \dots ..\left(i\right)$

Now, if this combination is immersed in liquid, then

$\frac{1}{F\prime }=\left({ }_1{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \dots ..\left(ii\right)$

$\frac{F\prime }{f}=\frac{({ }_a{\mu }_g-1)}{({ }_1{\mu }_g-1)}=\frac{\left(\left(1.5\right)-1\right) }{\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}$

$or \frac{F\prime }{f}=\frac{0.5}{\left(\frac{9}{8}-1\right)}=0.5\times 8$

$\therefore \mathrm{F}\text{'}=0.5\times 8\times 10=40 \mathrm{cm}$$\therefore \mathrm{F}\text{'}=0.5\times 8\times 10=40 \mathrm{cm}$