Ncert Solutions Physics Class 12th

9.12 Focal length of the objective lens, $f_o$ = 8 mm = 0.8 cm

Focal length of the eyepiece, $f_e$ = 2.5 cm

Object distance for the objective lens, $u_o$ = - 9.0 mm = -0.9 cm

Least distance of distant vision, d = 25 cm

Image distance of the eyepiece, $v_e$ = -d = -25 cm

Object distance of the eyepiece = $u_e$

Using the lens formula, we can obtain the value of $u_e$ as:

$\frac{1}{v_e}$ - $\frac{1}{u_e}$ = $\frac{1}{f_e}$ or $\frac{1}{u_e}$ = $\frac{1}{v_e}-\frac{1}{f_e}$ = $\frac{1}{-25}-\frac{1}{2.5}$

$u_e=-2.27cm$

Using the lens formula, we can obtain the value of $v_o$ as:

$\frac{1}{v_o}$ - $\frac{1}{u_o}$ = $\frac{1}{f_o}$ or $\frac{1}{v_o}$ = $\frac{1}{u_o}+\frac{1}{f_o}$ = $\frac{1}{-0.9}+\frac{1}{0.8}$

$v_o=7.2cm$

The distance between the objective lens and the eyepiece = $\left|u_e\right|$ + $v_o$ = 2.27 + 7.2 = 9.47 cm

The magnifying power of microscope is calculated as:

m = $\frac{v_o}{\left|u_o\right|}$ ( 1 + $\frac{d}{f_e\mathrm{}}$ ) = $\frac{7.2}{0.9}$ ( 1 + $\frac{25}{2.5\mathrm{}}$ ) = 88

Hence the magnification of the microscope is 88.

9.11 Focal length of the objective lens, $f_1$ = 2.0 cm

Focal length of the eyepiece, $f_2$ = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, d' = 25 cm

Hence, image distance for the eyepiece, $v_2$ = - 25 cm

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{-25}-\frac{1}{6.25}$

$u_2=-5cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 5 = 10 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{10}-\frac{1}{2}$

$u_1=-2.5cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.5 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}$ ( 1 + $\frac{d\text{'}}{f_2\mathrm{}}$ ) = $\frac{10}{2.5}$ ( 1 + $\frac{25}{6.25}$ ) = 20

Hence the magnifying power of the microscope is 20

The final image is formed at infinity.

Hence, $v_2$ = $\infty$

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{\infty }-\frac{1}{6.25}$

$u_2=-6.25cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 6.25 = 8.75 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{8.75}-\frac{1}{2}$

$u_1=-2.59cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.59 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}*$ $\frac{d\text{'}}{\left|u_2\right|}$ = $\frac{8.75}{2.59}*\frac{25}{6.25}$ = 13.51

Hence the magnifying power of the microscope is 13.51.

9.9 Size of the object, $h_1$ = 3 cm

Object distance, u = - 14 cm

Focal length of the concave lens, f = - 21 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{-14}$ = $-\frac{1}{21}$

$\frac{1}{v}$ = $-\frac{1}{14}-\frac{1}{21}$ or v = $-\frac{42}{5}$ = - 8.4 cm

Hence, the image is formed on the other side of the lens, 8.4 cm away from the lens. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

m = $\frac{Imageheight\left(h_2\right)}{Objectheight\left(h_1\right)}$ = $\frac{v}{u}$

$\frac{h_2}{h_1}$ = $\frac{-8.4}{-14}$ or $h_2$ = 0.6 $*3$ = 1.8 cm

If the object is moved further away from the lens, then the virtual image will move towards the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

9.6 The angle of minimum deviation, ${\delta }_m$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $=1.33$ , and the refractive index of prism material = $\mu \text{'}$

The angle of deviation is related to refractive index $\mu \text{'}$ is given as

${\mu }^{\text{'}}=\frac{\sin ?\frac{(A+{\delta }_m)}{2}}{\sin ?\frac{A}{2}}$ = $\frac{\sin ?\frac{(60°+40°)}{2}}{\sin ?\frac{60°}{2}}$ = $\frac{\sin ?50°}{\sin ?30°}$ = 1.532

So the refractive index of prism material is 1.532

Since the prism is placed in water, let ${\delta }_m^{\text{'}}$ be the new angle of minimum deviation.

The refractive index of glass with respect to water is given by the relation:

${\mu }_g^w$ = $\frac{{\mu }^{\text{'}}}{\mu }$ = $\frac{\sin ?\frac{(A+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{60°}{2}}$

1.152 $*\sin ?30°$ = $\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}$

$\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}={\sin }^{-1}?0.576$ = 35.2 $°$

${\delta }_m^{\text{'}}=2*$ 35.2 $°$ - 60 $°$ = 10.33 $°$

Hence the new minimum angle of deviation is 10.33$°$