Class 12th

9.4 In figure (a) – Glass-Air interface:

Angle of incidence, i = 60 $°$ , Angle of refraction, r = 35 $°$

The relative refractive index of glass with respect to air is given by Snell's law as:

${\mu }_g^a$ = $\frac{\sin ?i}{\sin ?r}$ = $\frac{\sin ?60°}{\sin ?35°}$ = 1.51 ………(1)

In figure (b) – Air - Water interface:

Angle of incidence, i = 60 $°$ , Angle of refraction, r = 47 $°$

The relative refractive index of water with respect to air is given by Snell's law as:

${\mu }_w^g$ = $\frac{\sin ?i}{\sin ?r}$ = $\frac{\sin ?60°}{\sin ?47°}$ = 1.18 ………(2)

Using equation (1) and (2), the relative refractive index of glass with respect to water can be obtained as

${\mu }_g^w$ = $\frac{{\mu }_g^a}{{\mu }_w^g}$ = $\frac{1.51}{1.18}$ = 1.28

In figure (c) – Water - Glass interface:

Angle of incidence, i = 45 $°$ , Angle of refraction = r

The relative refractive index of glass with respect to water is given by Snell's law as:

${\mu }_g^w$ = $\frac{\sin ?i}{\sin ?r}$ = $\frac{\sin ?45°}{\sin ?\mathrm{r}}$ = 1.28

sin r = $\frac{\sin ?45°}{1.28}$

r = 33.53^$°$

Hence, the angle of refraction at the water – glass interface is 33.53^$°$

9.2 This is an NCERT question from the class 12th Physics chapter 9 called 'Ray Optics and Optical instruments'. We will be explaining this question step-by-step for the convinience of students:

• Height of the needle, $h_1$ = 4.5 cm
• Object distance, u = - 12 cm
• Focal length of the convex mirror, f = 15 cm
• Image distance = v

The value of v can be obtained using the mirror formula

$\frac{1}{u}$ + $\frac{1}{v}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}$ - $\frac{1}{u}$ = $\frac{1}{15}$ - $\frac{1}{-12}$ = $\frac{1}{15}$ + $\frac{1}{12}$ = $\frac{9}{60}$

v = $\frac{60}{9}$ = 6.7 cm

Hence, the image of the needle is 6.7 cm away from the mirror. Also it is on the other side of the mirror.

The image size ( $h_2)$ is given by the magnification formula (m) as:

m = $\frac{h_2}{h_1}$ = $-\frac{v}{u}$ = $-\frac{6.7}{-12}$ = $\frac{6.7}{12}$

$h_2=\frac{6.7}{12}*4.5$ = 2.5 cm

Hence m = $\frac{2.5}{4.5}$ = 0.56

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual and diminished. If the needle is moved farther from the mirror, the image will also move away from the mirror and the size of the image will reduce gradually.

While this is one of the questions from the chapter, we have an NCERT excercise on Ray optics and optical instruments. Students who are currently in class 12th and plan on to take entrance exams, must go through this excercise for understanding how they can solve each question in proper steps.

9.1 Size of the candle, h = 2.5 cm

Let the image size be = h'

Object distance, u = $-$ 27 cm

Radius of curvature of the concave mirror, R = $-$ 36 cm

Focal length of the concave mirror, f = $\frac{R}{2}$ = $-$ 18 cm

Image distance = v

The image distance can be obtained by using mirror formula: $\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{1}{f}$ $-$ $\frac{1}{u}=\frac{1}{-18}$ $-\frac{1}{-27}$ = $-\frac{1}{54}$

v = −54 cm

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

m = $\frac{h\text{'}}{h}$ = $-\frac{v}{u}$

h' = $-\frac{v}{u}$ $*h$ = $-$ $\frac{-54}{-27}$ $*2.5$ = - 5 cm

The height of the candle's image is 5 cm. The negative sign indicates that the image is inverted and virtual. If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

1.8 (a) The situation is represented in the following figure.

O is the midpoint of AB, hence OA = OB = 10 cm = 10$*{10}^{-2}$ $m^2$

The electric field at O caused by the charge at A is given by

 $E_1$ =$\left|\frac{4\pi {?}_0}{}\mathrm{}*\frac{q_A}{r^2}\right|$ along OB and the electric field at O caused by the charge at B is given by,

 $E_2$ =$\left|\frac{1}{4\pi {?}_0}\mathrm{}*\frac{q_B}{r^2}\right|$along OB

where ${?}_0$= Permittivity of free space = 8.854 $*{10}^{-12}$
 $C^2{N}^{-1}$ $m^{-2}$

Net electric field at point O, E = $E_1+E_2$ = $\frac{1}{4\pi {?}_0{r}^2}$$*\left|q_A\mathrm{}+\mathrm{}{q}_B\right|$

$\left|{}_{}\right|$ $\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}{*(10\mathrm{}*{10}^{-2})}^2}$ $*\left|3*{10}^{-6}\mathrm{}+\mathrm{}3*{10}^{-6}\right|$ 

=5.39 $*{10}^6$ N$C^{-1}$ along OB

(b) A test charge of amount 1.5$*{10}^{-9}$  C is placed at the midpoint O

So the force experienced by the test charge, F = q $*E$

= 1.5  5.39$*{10}^6$ 
 N = 8.088 $*{10}^{-3}$
 N

This force is directed along the line OA, this is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.