Physics

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$°$

Initial kinetic energy of the rocket = $\frac{1}{2}{mv}^2$

Initial potential energy of the rocket = $\frac{-GMm}{R}$

Total initial energy = $\frac{1}{2}{mv}^2-\frac{GMm}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8 $*\frac{1}{2}{mv}^2$ $-\frac{GMm}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

Total energy of the rocket at height h = $-\frac{GMm}{R+h}$

Applying the law of conservation of energy for the rocket, we can write:

0.4 $*{mv}^2$ $-\frac{GMm}{R}$ = $-\frac{GMm}{R+h}$

0.4 $v^2$ = GM( $\frac{1}{R}$ $-\frac{1}{R+h})$

0.4 $v^2$ = GM( $\frac{R+h-R}{R(R+h)}$ ) = $\frac{GMh}{R(R+h)}$

$\frac{R+h}{h}$ = $\frac{GM}{{0.4Rv}^2}$

$\frac{R}{h}$ + 1 = $\frac{GM}{{0.4Rv}^2}$ = $\frac{6.4*{10}^{23}*6.67*{10}^{-11}}{0.4*3395*{10}^3*{2000}^2}$

h = 495 km

Mass of the spaceship, $m_s$ = 1000 kg

Mass of the Sun, M = 2 * 10³⁰ kg

Mass of Mars, $m_m$ = 6.4*10²³ kg

Orbital radius of Mars, R = 2.28 *10⁸ km = 2.28 *10¹¹ m

Radius of Mars, r = 3395 km = 3.395 $*{10}^{6}m$

Universal Gravitational constant, G = 6.67*10^-11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun = $\frac{-GM{m}_s}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars= $\frac{-G{m}_s{m}_m}{r}$

Since the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zero

Total energy of the spaceship = $\frac{-GM{m}_s}{R}$ $-\frac{-G{m}_s{m}_m}{r}$ = $-G{m}_s(\frac{M}{R}$ + $\frac{m_m}{r}$ )