If an object is placed at 25 cm from the correcting lens, it should produce the virtual image at 50 cm. Thus, u = - 25 cm ; v = - 50 cm
\(\frac{1}{\mathrm{f}}  =\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}  \)
\(=\frac{1}{-50}-\frac{1}{-25}.  \)
\(\frac{1}{\mathrm{f}}  =\frac{1}{23}-\frac{1}{50}=\frac{2-1}{50}  \)
\(\frac{1}{\mathrm{f}}  =\frac{1}{50}  \)
\(\mathrm{f}  =50 \mathrm{~cm}  \)
\(\text {Power } =\frac{1}{\mathrm{f}}=+\frac{1}{0.5}  \)
\(\mathbf{P}  =+2.0 \mathrm{D}\)
The unaided eye can see a maximum distance of 500 cm suppose the maximum distance for clear vision is d when the lens is used. Then the object at a distance d is imaged by the lens at 500 cm.
We have, \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\) 
\(-\frac{1}{500}-\frac{1}{\mathrm{d}} =\frac{1}{50}\) 
\(\frac{1}{500}+\frac{1}{d}=-\frac{1}{50}\) 
\(\frac{1}{d}=-\frac{1}{50}-\frac{1}{500}=-\frac{11}{500}\) 
\(d=\frac{500}{11}=-45.45 cm\) 
∴ Maximum distance, d = 45.45 cm

i_p = 60°
We know, r_p = 90° - i_p
The angle of refraction \(r_p=90^{\circ}-60^{\circ}=30^{\circ}\)
By Brewster's law, \(n^i=\tan i_p\)
\(\mathrm{n}=\tan 60^{\circ}=\sqrt{3}\)
We know, \(\mathrm{n}=\frac{1}{\sin \mathrm{c}}\) where, c is the critical angle
\(\therefore \sin c  =\frac{1}{n}=\frac{1}{\sqrt{3}}  \)
\(=\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{1.732}{3} \)
sin c  = 0.577 
c = 35^o 15'

I = 3 Wm^-2,
I₀ = 32Wm^-2,
Consider the following figure

\(\mathrm{I}=32 \mathrm{Wm}^{-2}  \)
\(\mathrm{I}_1=\frac{\mathrm{I}}{2}=16 \mathrm{Wm}^{-2} \)
\(\mathrm{I}_2=\mathrm{I}_1 \cos ^2 \theta_1 \quad \theta_1 \text { is the angle between the optic axes of } \mathrm{P}_1 \text { and } \mathrm{P}_3  \)
\(\mathrm{I}_2=16 \cos ^2 \theta_1  \)
\(\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \theta_2 \)
θ₂ is the angle between the optic axes of P₂ and P₃
\(I_3=\left(16 \cos ^2 \theta_1\right) \cos ^2 \theta_2  \)
\(\text { i.e., } \frac{3}{16}=\cos ^2 \theta_1 \cdot \cos ^2 \theta_2 \)
\(If \ \theta_1=30^{\circ} ; \quad \theta_2=60^{\circ} (Let \ us \ assume) \)
\(\frac{3}{16}=\left(\frac{\sqrt{3}}{2}\right)^2 \times\left(\frac{1}{2}\right)^2=\frac{3}{16}  \)
\(\text { LHS }=\text { RHS }\)
Therefore, the angle between the axes of P₁ and P₃ is 30^o.

The following figure clearly depicts the situation given

Let I be the intensity of unpolarised light which falls on polaroid P₁
Then, I =\(\frac{I}{2}\)
Since the angle between the optic axes of P₁ and P₂ is 90°, no light is emerging from polaroid P₂. After introducing the polaroid P₃ between P₁ and P₂ let the angle between the optic axes of P₁ and P₃ be θ₁ = 30°. If θ₁ = 30, the angle between the optic axes of P₃ and P₂, will be 60°.
For this values i.e., θ₁ = 30° & θ₂ = 60°
I₂ = I₁ cos² ∝ θ₁                    I₂ = I₁ cos² ∝ θ₁    
\(=\frac{I}{2} \cos ^{2} 30^o\)                 and I₃ = I₂ cos² 60^o  
\(=\frac{I}{2}\times\frac{3}{4}\)        \(=\frac{3I}{8}\times\frac{1}{4}\) 
\(I_2=\frac{3I}{8} .......(1) \quad \quad\quad I_3=\frac{3I}{32} .......(2)\) 
Instead, if θ₁ =  45° and θ₂ = 45°
\(NowI_2=\frac{I}{2}\times cos^245^o \quad and\quad I_3=\frac{I}{4}\times cos^245^o\)
\(=\frac{I}{2}\times\frac{1}{2} \quad \quad I_3=\frac{I}{8} ......(8)\)
\(​​​​​​​I_2=\frac{I}{4}  ......(3)\)
Compare the results (2) and (4)
I₃ value for 45° is greater.
Therefore, we conclude that if the angle between the axes of first and newly introduced polaroid is 45°, the emerging light from the polaroid will be greater.

λ = 5000 Å = 0.5 x 10^-6 m, 
a = 2.5 μm = 2.5 x 10^-6 m
We know, a sin θ = nλ
For maximum order of diffraction, sin θ = 1
Therefore, 2.5 x 10^-6 = n x 0.5 x 10^-6
\(n=\frac{2.5}{0.5}=5\) 
Maximum order of diffraction = 5.

(i) Its cost is very high due to scarity of large and flawless calcite crystals
(ii) Due to extraordinary ray passing obliquely through it the emergent ray is always displaced a little to one side.
(iii) The effective field of view is quite limited
(iv) Light emerging out of it is not uniformly plane polarised.

(i) Similar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that become fainter while moving away from the center as shown in Figure. These are known as Airy's discs
\(asin\theta =1.22\lambda \) 

(ii) Here, the numerical value 1.22 comes for central "maximum formed by circular apertures.
For small angles, sin θ = θ
\(a\theta =1.22\lambda \) 
Rewriting further,
\(\theta =\cfrac { 1.22\lambda }{ a } \ and \cfrac { { r }_{ o } }{ f } =\cfrac { 1.22\lambda }{ a } \)
\({ r }_{ o }=\cfrac { 1.22\lambda f }{ a } \)

(i) For points of maxima, the slit is to be divided into an odd number of equal parts so that one part remains un-canceled making the point P appear bright.
The condition for the first maximum is,
\(\cfrac { a }{ 3 } sin\theta =\cfrac { \lambda }{ 2 } \) (or) \(asin\theta =\cfrac { 3\lambda }{ 2 } \) 
The condition for the second maximum is
\(\cfrac { a }{ 5 } sin\theta =\cfrac { \lambda }{ 2 } \) (or) \(asin\theta =\cfrac { 5\lambda }{ 2 } \) 
The condition for the third maximum is,
\(\cfrac { a }{ 7 } sin\theta =\cfrac { \lambda }{ 2 } \) (or) \( asin\theta =\cfrac { 7\lambda }{ 2 } \)
In the same way, the condition for n^th maximum is 
\(asin\theta =\left( 2n+1 \right) \cfrac { \lambda }{ 2 } \) (n^th maximum)
where, n = 0, 1, 2, 3, ... , is the order of diffraction maximum.
(ii) The central maximum is called the order maximum. The points of the maximum intensity lie nearly midway between the successive minima.

(i) The distance D between the screen and douhle slit should be as large as possibie.
(ii) The wavelength \(\lambda\) of light used must be as long as possible.
(iii) The distance d between the two slits must be as small as possible.