Here, f₀ = 100mm = 1cm, f_e = 30mm = 3cm
u₀ = -1.2cm, v_e = -0.25m = -25cm
M = ? L = ?
From \({1\over v_0}-{1\over u_0}={1\over f_0.}\)
\({1\over v_0}={1\over f_0}+{1\over u_0}={1\over1}-{1\over1.2}={0.2\over1.2}={1\over6}\)
\(v_0=6cm\)
From \({1\over v_e}-{1\over u_e}={1\over f_e}\)
\({1\over u_e}={1\over v_e}-{1\over f_e}={1\over -25}-{1\over 3}={-28\over 75}\)
\(u_e={-75\over28}=-2.7cm\)
\(M={v_0\over -u_0}\left(1+{d\over f_e}\right)={6\over-1.2}\left(1+{25\over3}\right)\)=-46.7
L = v₀ + |u_e| = 6 + 2.7 = 8.7cm

\(Here \ { \mu }_{ 1 }=1.33,d=1 \ mm={ 10 }^{ -3 }m,\)
\(D=1.33m,\ \lambda =6300\mathring { A } =6.3\times{ 10 }^{ -7 }m\)
\(When \ the \ apparatus \ is \ immersed \ in \ the \ \lambda \)
\( changes \ to\lambda '=\frac { \lambda }{ { \mu }_{ l } } \)
\((i) \therefore =Fringe\quad width\)
\( \beta =\frac { \lambda 'D }{ d } =\frac { \lambda D }{ { \mu }_{ l }d }\)
\(=\frac { 6.33\times{ 10 }^{ -7 }\times1.33 }{ 1.33\times{ 10 }^{ -3 } } \)
\(or \ \beta =6.3\times{ 10 }^{ -4 }m=0.63m\)
\((ii)Displacement \ of \ fringes\)
\( \triangle y=\frac { \beta }{ \lambda ' } (\frac { { \mu }_{ g } }{ { \mu }_{ l } } -1)t\)
\(or \ \triangle y=-\frac { \beta }{ \lambda ' } ({ \frac { { \mu }_{ g }{ \mu }_{ l } }{ { \mu }_{ r } } ) }^{ t }\)
or
\( \triangle y=\frac { \beta }{ \lambda } ({ \mu }_{ g }-{ \mu }_{ l })t\quad d [\because \lambda '{ \mu }_{ l }=\lambda ]\)
\(or \ \ t=\frac { \triangle y\lambda }{ \beta ({ \mu }_{ g }-{ \mu }_{ l }) } \)
\(But,\ we \ know \ that \ \triangle y \ is \ the \ separation \ between \ adjacent \ dark \ and \ bright \ fringe\)
\(i.e. \triangle y=\frac { \beta }{ 2 } \)
\( t=\frac { \frac { \beta }{ 2 } .\lambda }{ \beta ({ \mu }_{ g }-{ \mu }_{ l }) } =\frac { \lambda }{ 2({ \mu }_{ g }-{ \mu }_{ l }) } \)
\(or\ t=\frac { 6.3\times{ 10 }^{ -7 } }{ 2(1.53-1.33) } =\frac { 6.3\times{ 10 }^{ -7 } }{ 2\times0.20 } \)
\( =1.575\times{ 10 }^{ -6 }=1.575\mu m\)

Given \({ v }_{ 0 }=3.3*{ 10 }^{ 14 }\) Hz

           \({ v }=8.2*{ 10 }^{ 14 }\)

Since  \({ eV }_{ 0 }=hv-h{ v }_{ 0 }\)

So         \({ v }_{ 0 }=\frac { h }{ e } (v-{ v }_{ 0 })\)

                   =\(\frac { 6.62*{ 10 }^{ -34 } }{ 1.6*{ 10 }^{ -19 } } *8.2*{ 10 }^{ 14 }\)

                                    \(-3.3*{ 10 }^{ 14 }\)

                    = \(4.14*{ 10 }^{ -15 }*4.9*{ 10 }^{ 14 }\)

                     =2.02 V

                     =2.0 V.

(a) The values noticed in Shyam are:
(i) High degree of general awareness.
(ii) Concern for his friend.
(iii) Helping and caring nature.
(b) The threshold wavelength, \({ \lambda }_{ 0 }=\frac { hc }{ W } \)
For lithium, \({ \lambda }_{ 0 }=\frac { 12375 }{ 2.3 } \overset { 0 }{ A } =5380\overset { 0 }{ A } \)
For copper, \({ \lambda }_{ 0 }=\frac { 12375 }{ 4 } \overset { 0 }{ A } =3094\overset { 0 }{ A } \) 
The wavelength 5380\(\overset { 0 }{ A } \) lies in visible region, thus lithium will be useful for photoelectric cell.

Here, \(A=60°,\ \mu =1.5,\)
\(i_{ 1 }=i_{ 2 }=\frac { 3 }{ 4 } \times 60°=45°,\delta =?\)
As \(A+\delta =i_{ 1 }+i_{ 2 }\)
\(\therefore \) \(60°+\delta =45°+45°,\) or \(\delta =90°-60°=30°\)

\(Here,\lambda =5500\overset { \circ }{ A } =5500\times { 10 }^{ -10 }m\)
Rate of solar energy reaching earth from the sun
\(=2 \ cal{ cm }^{ -2 }{ min }^{ -1 }=2\times 4.2J{ cm }^{ -2 }{ min }^{ -1 }\)
\(=2\times 4.2\times 60J{ cm }^{ -2 }{ hour }^{ -1 }\)
Energy of photon received on earth
 \(E=\frac { hc }{ \lambda } =\frac { \left( 6.6\times { 10 }^{ -34 } \right) \times \left( 3\times { 10 }^{ 8 } \right) }{ 5500\times { 10 }^{ -10 } } \)
\( =3.6\times { 10 }^{ -19 }J\)
If n number of photons is reaching the earth per cm² per hour, then total energy will be
\(=nE=n\times 3.6\times { 10 }^{ -19 }J{ cm }^{ -2 }{ h }^{ -1 }\)
\( \therefore n\times 3.6\times { 10 }^{ -19 }=2\times 4.2\times 60\)
\(or \ n=\frac { 2\times 4.2\times 60 }{ 3.6\times { 10 }^{ -19 } } =1.4\times { 10 }^{ 21 }\)

Wavelength of the monochromatic radiation, λ = 640.2 nm
= 640.2 x 10⁻⁹ m
Stopping potential of the neon lamp, V₀ = 0.54 V
Charge on an electron, e = 1.6 x 10⁻¹⁹ C
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Let ϕ₀ be the work function and ν be the frequency of emitted light.
We have the photo-energy relation from the photoelectric effect as:
eV₀ = hν − ϕ₀
\(\phi_{0}=\frac{\mathrm{hc}}{\lambda}-e V_{0}\)
\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{640.2 \times 10^{-9}}-1.6 \times 10^{-19} \times 0.54\)
\(=3.093 \times 10^{-19}-0.864 \times 10^{-19}\)
\(=2.229 \times 10^{-19} J\)
\(=\frac{2.229 \times 10^{-19}}{1.6 \times 10^{-19}}=1.39 \mathrm{eV}\)
Wavelength of the radiation emitted from an iron source, λ' = 427.2 nm
= 427.2 x 10⁻⁹ m
Let V′₀be the new stopping potential. Hence, photo-energy is given as:
\(e V_{0}^{\prime}=h \frac{c}{\lambda} \prime-\phi o_{0}\)
\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{427.2 \times 10^{-9}}-2.229 \times 10^{-19}\)
\(=4.63 \times 10^{-19}-2.229 \times 10^{-19}\)
\(=2.401 \times 10^{-19} J\)
\(=\frac{2.401 \times 10^{-19}}{1.6 \times 10^{-19}}=1.5 \mathrm{eV}\)
Hence, the new stopping potential is 1.50 eV.

Energy of one photon 
\(E=\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 6\times { 10 }^{ -7 } } \)  
\( \simeq 3.3\times { 10 }^{ -19 }J\)
E₁ = energy emitted by the source in one second = 66J
\(\therefore \) Number of photons emitted by the source in
\(1s=n=\frac { 66 }{ 3.3\times { 10 }^{ -19 } } =2\times { 10 }^{ 20 }\)
\(\therefore \) Total number of photons emitted by source in 2 minutes
\(=N=n\times 2\times 60\)
\( =2\times { 10 }^{ 20 }\times 120\)
\(=2.4\times { 10 }^{ 22 }photons\)

A photoelectric cell is called an electric eye as the photoelectric current set up in the photoelectric cell corresponding to incident light provides the information about the objects as has been seen by our eye in the presence of light.