Physics Ncert Solutions Class 12th

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.

1.6

In the adjoining figure ABCD is a square with sides AB = BC = CD = DA = 10 cm

Diagonals, AC = BD  = $\sqrt[]{{10}^2+{10}^2}$ = $10\sqrt[]{2}$cm

AO = OC = DO = BO = $5\sqrt[]{2}$ cm

At the centre of the square ABCD, O, a charge of 1 $\mu Cisplaced.$$\text{exception 25:}$

The force of repulsion between the charges placed at A and at O is equal in magnitude but opposite in direction between the charges placed at point C and centre O. Similarly ,the force of attraction between the charges placed at B & O and D & O will be equal in magnitude but opposite in direction. These charges will cancel each other.

Hence, the net charge at centre O will be zero.

Here is a deeper explanation. 

What you just saw is the use of the superposition principle. The total force on the 1 μC charge at the centre (O) is the vector sum of the forces from each of the four corner charges. 

Pairs can break down the analysis. 

• Forces from qA and qC: The force exerted by qA (+2 μC) on the central charge is repulsive. The force from qC (+2 μC) is also repulsive. Now, since qA and qC are equal and are at an equal distance from the centre, the forces they exert are equal in magnitude. But they point in opposite directions along the diagonal AC. So, they cancel each other out.

• Forces from qB and qD: Similarly, the force exerted by qB (–5 μC) is attractive, and the force from qD (–5 μC) is also attractive. Because these charges are equal and equidistant from the centre, the forces are equal in magnitude but opposite in direction along the diagonal BD. That causes them to cancel out as well.

As both pairs of forces nullify each other, the net electrostatic force on the central charge is zero. Read more about Coulomb's Law