(1)

For thin lenses in contact, then the powers of each lens add up.

$P_t=P_1+P_2+P_3+P_4=4P$

Given linear magnification of each lens $m$ is

The overall magnification of the combined lens system is

$M=m\times m\times m\times m=m^4 \text{(dimensionless).}$

Hence the combination has power 4P and magnification $m^4$.

4P and $m^4$ -- follows from the additive power law for co-axial lenses and the multiplicative rule for successive magnifications.

(4)

Since, the given lens is made up of three different materials so there are three refractive indices, $\mu$ and hence, three images will be formed.

(3)

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The same size image is formed when object is positioned at centre of curvature. Using geometric configuration, $x=2f$ or $f=x/2$

(2)

Let the focal length of the convex lens be $\mathit{\text{'f'}}$.

When two lenses are placed in contact with each other, their power will be added.

$\therefore$    $P_{\text{combination}}=P_{\text{convex}}+P_{\text{concave}}$

$\frac{1}{f_{\text{combination}}}=\frac{1}{f_{\text{convex}}}+\frac{1}{f_{\text{concave}}}=\frac{1}{f}+\frac{1}{(-f)}=0$

$\therefore$   $f_{\text{combination}}=\frac{1}{0}=\infty$

Hence, $f_{\text{combination}}$ will be infinite.

(1)

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From lens maker's formula,

$\frac{1}{f_1}=(1.6-1)(\frac{{\displaystyle 1}}{{\displaystyle \infty }}-\frac{{\displaystyle 1}}{{\displaystyle 20}})=-\frac{0.6}{20}$

$f_1=\frac{-100}{3}\text{cm} ; \frac{1}{f_2}=(1.5-1)(\frac{{\displaystyle 1}}{{\displaystyle 20}}+\frac{{\displaystyle 1}}{{\displaystyle 20}})$

$f_2=20\text{cm}$

$f_3=f_1=\frac{-100}{3}\text{cm} ; \frac{1}{f_{\text{net}}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{-3}{100}+\frac{1}{20}-\frac{3}{100}=\frac{-1}{100}$

$f_{\text{net}}=-100\text{cm}$

(3)

Given, refractive index of lens = 1.5

For biconvex lens, $R_1=+R, R_2=-R$

$\therefore$$R_1=+20\text{cm} \text{and} R_2=-20\text{cm}$

According to lens maker’s formula,

$\frac{1}{f}=(\mu -1)(\frac{{\displaystyle 1}}{{\displaystyle R_1}}-\frac{{\displaystyle 1}}{{\displaystyle R_2}})$  $\therefore$$\frac{1}{f}=(1.5-1)(\frac{1}{20}-\frac{1}{(-20)})=0.5(\frac{1}{20}+\frac{1}{20})$

or     $\frac{1}{f}=\frac{1}{20} \Rightarrow f=20\text{cm}$

Power of lens (in dioptre),  $P=\frac{100}{\text{focal length }f \left(\text{in cm}\right)}$

$\therefore$    $P=\frac{100}{20}=+5D$

(3)

Given : $f_1=20\text{cm}, f_2=-5\text{cm}$

Equivalent focal length, $f=\infty$

(As the rays are parallel)

By using Newton’s displacement formula,

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2} \Rightarrow \frac{1}{\infty }=\frac{1}{20}-\frac{1}{5}-\frac{d}{20\times (-5)}$

$0=\frac{5-20+d}{20\times 5} \Rightarrow d=15\text{cm}$

(3)

According to lens maker’s formula

$\frac{1}{f}=(\mu -1)(\frac{{\displaystyle 1}}{{\displaystyle R_1}}-\frac{{\displaystyle 1}}{{\displaystyle R_2}})$

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$\frac{1}{f}=(\mu -1)(\frac{{\displaystyle 1}}{{\displaystyle R}}-\frac{{\displaystyle 1}}{{\displaystyle -R}})=(1.5-1)\left(\frac{{\displaystyle 2}}{{\displaystyle R}}\right)=\frac{1}{R}$

Two similar equi-convex lenses of focal length $f$ each are held in contact with each other.

The focal length $F_1$ of the combination is given by

               $\frac{1}{F_1}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f} ; F_1=\frac{f}{2}=\frac{R}{2}$                                ...(i)

For glycerin in between lenses, there are three lenses, one concave and two convex.

Focal length of the concave lens is given by

$\frac{1}{f\text{'}}=(1.5-1)\left(\frac{{\displaystyle -2}}{{\displaystyle R}}\right)=-\frac{1}{R}$

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Now, equivalent focal length of the combination is,

$\frac{1}{F_2}=\frac{1}{f}+\frac{1}{f\text{'}}+\frac{1}{f} ; \frac{1}{F_2}=\frac{1}{R}-\frac{1}{R}+\frac{1}{R}=\frac{1}{R}$

$F_2=R$                                                                                     ...(ii)

Dividing equation (i) by (ii), we get $\frac{F_1}{F_2}=\frac{1}{2}$

(4)

When an equiconvex lens is cut into two symmetrical halves along the principal axis, then there will be no change in focal length of the lens.

$\therefore$   $\text{Power of lens,} P=\frac{1}{f}$

So, the power of each part will be $P.$

(4)

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Here, ${\mu }_g=\frac{3}{2}, {\mu }_w=\frac{4}{3}$

Focal length $\left(f\right)$ of glass convex lens is given by

         $\frac{1}{f}=({\mu }_g-1)\left(\frac{{\displaystyle 2}}{{\displaystyle R}}\right)$

or    $\frac{1}{f}=(\frac{{\displaystyle 3}}{{\displaystyle 2}}-1)\frac{2}{R}=\frac{1}{R} \text{or} f=R$                                  ...(i)

Focal length $(f\text{'})$ of water filled concave lens is given by

          $\frac{1}{f\text{'}}=({\mu }_w-1)(-\frac{{\displaystyle 2}}{{\displaystyle R}}) \text{or} \frac{1}{f\text{'}}=(\frac{{\displaystyle 4}}{{\displaystyle 3}}-1)(-\frac{{\displaystyle 2}}{{\displaystyle R}})$

                     $=-\frac{2}{3R}=-\frac{2}{3f} \text{[Using eqn. (i)]}$

Equivalent focal length $\left(f_{\text{eq}}\right)$ of lens system

         $\frac{1}{f_{\text{eq}}}=\frac{1}{f}-\frac{2}{3f}+\frac{1}{f}=\frac{3-2+3}{3f}=\frac{4}{3f} \therefore f_{\text{eq}}=\frac{3f}{4}$