Class 12th

This is a short answer type question as classified in NCERT Exemplar

apparent depth = real depth / refractive index

Since, the image formed by Medium1, O₂ act as an object for Medium2.

If seen from µ₃ ,the apparent depth is O₂.

Similarly, the image formed by Medium2, O₂ act as an object for Medium3

O₂₌$\frac{{\mu }_3}{{\mu }_2}$ ( $\frac{h}{3}+O_1$ )

= $\frac{{\mu }_3}{{\mu }_2}$ ( $\frac{h}{3}+\frac{{\mu }_2}{{\mu }_1}\frac{h}{3})$

Seen from outside the apparent height is

O₃ = $\frac{1}{{\mu }_3}$ (  $\frac{h}{3}+O_2)$ = $\frac{1}{{\mu }_3}\left(\frac{h}{3}+\frac{h}{3}\left(\frac{{\mu }_3}{{\mu }_2}+\frac{{\mu }_3}{{\mu }_1}\right)\right)$

= $\frac{h}{3}(\frac{1}{{\mu }_1}+\frac{1}{{\mu }_2}+\frac{1}{{\mu }_3}$ )

This is the expression for required depth. 

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill x+4\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x+4\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x\hfill & \hfill x+4\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill 3x+4\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill 3x+4\hfill & \hfill x+4\hfill & \hfill x\hfill \\ \hfill 3x+4\hfill & \hfill x\hfill & \hfill x+4\hfill \end{array}\right|\\ Taking\left(3x+4\right)commonfrom{C}_1\\ =\left(3x+4\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill 1\hfill & \hfill x+4\hfill & \hfill x\hfill \\ \hfill 1\hfill & \hfill x\hfill & \hfill x+4\hfill \end{array}\right|\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ =\left(3x+4\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill -4\hfill \\ \hfill 1\hfill & \hfill x\hfill & \hfill x+4\hfill \end{array}\right|\\ Expandingalong{C}_1=\left(3x+4\right)\left[1\left|\begin{array}{cc}\hfill -4\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -4\hfill \end{array}\right|\right]\\ =\left(3x+4\right)\left(16-0\right)=16\left(3x+4\right)\end{array}$

This is a short answer type question as classified in NCERT Exemplar

no reversibility of lens maker formula is not possible.

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill 3x\hfill & \hfill -x+y\hfill & \hfill -x+z\hfill \\ \hfill x-y\hfill & \hfill 3y\hfill & \hfill z-y\hfill \\ \hfill x-z\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill x+y+z\hfill & \hfill -x+y\hfill & \hfill -x+z\hfill \\ \hfill x+y+z\hfill & \hfill 3y\hfill & \hfill z-y\hfill \\ \hfill x+y+z\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ Taking\left(x+y+z\right)commonfrom{C}_1\\ =\left(x+y+z\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -x+y\hfill & \hfill -x+z\hfill \\ \hfill 1\hfill & \hfill 3y\hfill & \hfill z-y\hfill \\ \hfill 1\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ =\left(x+y+z\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -x-2y\hfill & \hfill -x+y\hfill \\ \hfill 0\hfill & \hfill 2y+z\hfill & \hfill -y-2z\hfill \\ \hfill 0\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ Expandingalong{C}_1=\left(x+y+z\right)\left[1\left|\begin{array}{cc}\hfill -x-2y\hfill & \hfill -x+y\hfill \\ \hfill 2y+z\hfill & \hfill -y-2z\hfill \end{array}\right|\right]\\ =\left(x+y+z\right)\left[\left(-x-2y\right)\left(-y-2z\right)-\left(2y+z\right)\left(-x+y\right)\right]\\ =\left(x+y+z\right)\left(xy+2zx+2y^2+4yz+2xy-2y^2+zx-zy\right)\\ =\left(x+y+z\right)\left(3xy+3zx+3yz\right)\\ =3\left(x+y+z\right)\left(xy+yz+zx\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill 0\hfill & \hfill x{y}^2\hfill & \hfill x{z}^2\hfill \\ \hfill x^{2}y\hfill & \hfill 0\hfill & \hfill y{z}^2\hfill \\ \hfill x^{2}z\hfill & \hfill z{y}^2\hfill & \hfill 0\hfill \end{array}\right|\\ Takingx{,}^2{y}^{2}and{z}^{2}commonfrom{C}_1,C_{2}and{C}_{3}respectively\\ =x^2{y}^2{z}^2\left|\begin{array}{ccc}\hfill 0\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill y\hfill & \hfill 0\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill z\hfill & \hfill 0\hfill \end{array}\right|\\ Expandingalong{R}_1=x^2{y}^2{z}^2\left[0\left|\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill 0\hfill \end{array}\right|-x\left|\begin{array}{cc}\hfill y\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill 0\hfill \end{array}\right|+x\left|\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill z\hfill & \hfill z\hfill \end{array}\right|\right]\\ =x^2{y}^2{z}^2\left[-x\left(0-yz\right)+x\left(yz-0\right)\right]\\ =x^2{y}^2{z}^2\left(xyz+xyz\right)=x^2{y}^2{z}^2\left(2xyz\right)=2x^3{y}^3{z}^3\end{array}$

This is a short answer type question as classified in NCERT Exemplar

m=v/u = D/f

So m = 25/10 = 2.5= 0.025m

P= 1/0.025= 40D

This is the required power of lens.

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill a+x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill a+y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill a+x+y+z\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill a+x+y+z\hfill & \hfill a+y\hfill & \hfill z\hfill \\ \hfill a+x+y+z\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|=\left(a+x+y+z\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill 1\hfill & \hfill a+y\hfill & \hfill z\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|\\ \left(Takinga+x+y+zcommon\right)\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ =\left(a+x+y+z\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill a\hfill & \hfill -a\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|\\ Expandingalong{C}_1=\left(a+x+y+z\right)\left|1\left(a^2-0\right)\right|=a^2\left(a+x+y+z\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- (i) Here it is given that, an electron absorbs 2 photons each of frequency ν then ν where, v′ is the frequency of emitted electron.

Given, E_max= hv- ${?}_0$

Now, maximum energy for emitted electrons is E_max= h2v- ${?}_0=2\mathrm{h}\mathrm{v}-{?}_0$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible

This is a short answer type question as classified in NCERT Exemplar

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

By lens makers formula 1/f= $(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

So f_br so focal length of blue is less than red light