λ = 5000 Å = 5000 x 10 ^-10 m; sin 30^o = 0.5; n = 1; a =?
Equation for diffraction minimum is,
asin θ = nλ
The central maximum is spread up to the first minimum. Hence, n = 1
Rewriting, \(a=\cfrac { \lambda }{ sin\theta } \) 
Substituting, \(a=\cfrac { 5000\times { 10 }^{ -10 } }{ 0.5 } \) 
a = 1 x 10^-6m = 0.001 x 10^-3 m = 0.001 mm.

Amplitudes, a₁ = 5, a₂ = 3
Resultant amplitude,
\(A=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+2{ a }_{ 1 }{ a }_{ 2 }cos\varphi } \) 
Resultant amplitude is maximum when,
\(\phi =0,cos0=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \) 
 \(\\ { A }_{ max }=\sqrt { \left( { a }_{ 1 }+{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5+3 \right) ^{ 2 } } =\sqrt { \left( 8 \right) ^{ 2 } } \) 
= 8 units
Resultant amplitude is minimum when
\(\phi =\pi,cos\pi=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \)
\({ A }_{ max }=\sqrt { \left( { a }_{ 1 }-{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5-3 \right) ^{ 2 } } =\sqrt { \left( 2 \right) ^{ 2 } } \)
= 2units
\(I\infty { A }^{ 2 }\) 
\(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( { A }_{ max } \right) ^{ 2 } }{ \left( { { A }_{ min } } \right) ^{ 2 } } \)
Substituting,
 \(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( 8 \right) ^{ 2 } }{ \left( 2 \right) ^{ 2 } } =\cfrac { 64 }{ 4 } 16\) (or)
\({ I }_{ max }:{ I }_{ min }=16:1\)

\(\mathrm{m}_{\alpha}=100 ; \mathrm{f}_{o}=0.5 \mathrm{~cm} ; \mathrm{f}_{\mathrm{e}}=? \)
\(\mathrm{~L}_{\alpha}=6.5 \mathrm{~cm}, \mathrm{D}=25 \mathrm{~cm} \)
When the image is formed at infinity,
\(\mathrm{m}_{\alpha}=\frac{\left(\mathrm{L}_{\alpha}-\mathrm{f}_{0}-\mathrm{f}_{\mathrm{c}}\right) \mathrm{D}}{\mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}} \)
\(100 =\left(\frac{6.5-0.5-f_{e}}{0.5 \times f_{e}}\right) \times 25 \)
\(=\left(\frac{6-f_{e}}{0.5 f_{e}}\right) \times 25 \)
\(100 \times 0.5 f_{e} =150-25 f_{e} \)
\(50 f_{e} =150-25 f_{e} \)
\(75 f_{e} =150 \)
\(f_{e} =150 / 75=2 \mathrm{~cm} \)

From figure,
D = u + v
d = r - u
D + d = u + v + v - u
D + d = 2v

\(v=\frac{D+d}{2}\)
D - d = u + v - v + u = 2u
\(v=\frac{D-d}{2}\)
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(\frac{1}{f} =\frac{1}{\frac{D+d}{2}}-\frac{1}{\frac{D-d}{2}} \)
\(=\frac{2}{D+d}-\frac{2}{D-d} \)
\(=\frac{2[D-d+D+d]}{D^{2}-d^{2}} \)
\(=\frac{2 \times 2 D}{D^{2}-d^{2}} \)
\(\frac{1}{f} =\frac{4 D}{D^{2}-d^{2}} \quad \therefore f=\frac{D^{2}-d^{2}}{4 D} \)

The minimum distance the person could see clearly is, y = 75 cm.
The lens should have a focal length of,
\(f=\cfrac { y\times 25cm }{ y-25cm } \) 
\(f=\cfrac { 75cm\times 25cm }{ 75cm-25cm } =37.5cm\) 
It is a convex (or) converging lens.
The power of the lens is,
\(P=\cfrac { 1 }{ 0.375m } =2.67\ D\)

The maximum distance the person could see is, x = 1.8 m.
The lens should have a focal length of,
f = –x m = –1.8 m.
It is a concave (or) diverging lens.
The power of the lens is,
\(\\ \\ \\ \\ P=-\cfrac { 1 }{ 1.8m } =-0.56D\)

Brewster’s law, tan i_p = n
For glass, tani_p = 1.5 ; i_p = tan^-11.5 ; i_p= 56.3^o
For water, tani_p= 1.33; i_p= tan^-1 = tan^-1 1.33; i_p = 53.1^o

a = 2.3 m; λ = 589 nm = 589 x 10^-9 m; θ = ?
The equation for angular resolution is,
\(\theta =\cfrac { 1.22\lambda }{ a } \) 
Substituting,
\(\theta =\cfrac { 1.22\times 589\times { 10 }^{ -9 } }{ 2.3 } =3.124\times { 10 }^{ -9 }\)
θ = 3.214 x 10^-7 rad (or) θ = 0.0011'
Note: The angular resolution of human eye is approximately, 3 x 10^-4 rad ≃ 1.03'.

λ = 500 nm = 500 x 10^-9 m; m = 4;
θ = 30°; number of lines per cm = ?
Equation for diffraction maximum in grating is, sin θ = Nm λ
Rewriting, \(N=\cfrac { sin\theta }{ m\lambda } \) 
Substituting,
\(N=\frac{0.5}{4 \times 500 \times 10^{-9}}\) 
= 2.5 x 10⁵ m^-1
= 2.5 x 10³ cm^-1

Let the amplitude be a.
The intensity is \(I\propto { 4a }^{ 2 }{ cos }^{ 2 }\left( \phi /2 \right) \) 
or \(I=4{ I }_{ 0 }{ cos }^{ 2 }\left( \phi /2 \right) \) 
Resultant intensity is maximum when,
\(\phi =0,cos=0=1,{ I }_{ max }\propto { 4a }^{ 2 }\) 
Resultant amplitude is minimum when,
\(\phi =\pi ,cos\left( \pi /2 \right) =0,{ I }_{ min }=0\) 
I_max : I_min = 4a^a : 0