A hollow prism contains air which does not cause dispersion. The faes AB and AC of the hollow prism behave like parallel sides of glass plates. The beam is laterally deviated at each of the two refracting faces. However, the rays of different colours emerge parallel to each other. So there is no dispersion.

1.33; 1.7 cm

f_e = 2 cm, f_o = 8 cm

(a) for a convex lens, f >>0 and for an object on left, u < 0. when the object is placed within the focus of a convex lens,
0 < |u| < f  or 0 < 1|u| > 1/f
1/v = 1/f+1/u=1/f-1/|u| < 0
i.e. v < 0 so a virtual image is formed on left.
Now as u<0 and v < 0, so 1/v = 1/f + 1/u
= - 1/ |v| = 1/f – 1/|u| or 1/|u| - 1/|v| = 1/f
As f > 0
1/|u| - 1/|v| > 0 or  1/|u| > 1/|v| or |u|< |u|
i.e |v| > |u| = |v/u| > 1
Hence image is enlarged
(b) For a concave lens f<0 and for an object on left, u<0
1/v = 1/f +1/u = 1/|f| - 1/|u|
 = - [1/|f| +1/|u|] < 0  for all u
i.e v < 0 for all values of u. hence a virtual image is formed on the left.
Also 1/|v| = 1/|f| + 1/|u|   1/|v| > 1/|u|
or |v|<|u  |m| = |v/u| < 1
i.e. the image is diminished in size

Magnification produced by any lens,
m = v/u = f / f + u
given m = ± N   ±N = f / f + u
or f + u = ± f / N or u = - f ± f / N
hence magnitude of object distances,
|u| = f ± f / N
given P = 1/f = + 2.5 D
f = 1/ 2.5 = 0.4 m = 40 cm
Also N = 4
|u| = 40 ± 40/4 = 40 ± 10 = 50 cm or 30 cm

(a) Polaroid sunglasses are preferred because they can be much more effective than coloured sunglasses in cutting off the harmful (UV) rays of the sun.
(b)

 

The diagram, given here, shows several fringes, due to double slit interference,' contained' in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.
[Note: The students may be awarded 1mark even if they just draw the diagram.] Two basic features:
(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.
(ii) Interference pattern is the superimposition of two waves slits originating from two narrow slits.The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.
(iii) For a single slit of width 'a' the first null of diffraction pattern occurs at an angle of \(\lambda\over a\). At the same angle of \(\lambda/a\), we get a maxima for two narrow slits separated by a distance

The reflected, completely plane polarised light makes an angle of 90° with the direction of corresponding refracted light.
Brewster's law According to Brewster, the polarising angle i_p, the reflected plane polarised light and refracted rays are perpendicular to each other, then
i_p + r = 90°
where, r = angle of refraction
r = 90°- i_p
\(\because\) Snell's law, \(\mu={{sin\ {i}_{p}}\over{sin\ r}}={{sin\ {i}_{p}}\over{sin(90°-{i}_{p})}}\)
\(\mu={{sin\ {i}_{p}}\over{cos \ {i}_{p}}}\Rightarrow\mu = tan\ {i}_{p}\)
This relation is known as Brewster's law.

The required graph would have the form as shown in figure below:

Using I2 = I1 cos² \(\theta\)
Given
I₁ = light transmitted by F₁
I₂ = light transmitted by P₂
I₃ = light transmitted by P₃
According to Malus law
I₃ = I1 cos² \(\beta \)                                    ...(i)
I₂ = I3 cos²\((\theta-\beta)\)                            ...(ii)
According to question
I₂=I₃
Substituting the value of I₂ and I₃ from Eq. (i) and Eq. (ii), we get
I₃cos²\((\theta-\beta)\) = I1cos²\(\beta\)
Substituting the value of I₃ from Eq. (i),
I₁cos²\(\beta\)cos²\((\theta-\beta)\) = I1 cos2 \(\beta\)
cos²\((\theta-\beta)\) = 1
\((\theta-\beta)\) = cos-1(1)
\((\theta-\beta)\) = 0
\(\theta=\beta\)                                                               ...(iii)
According to question I₁ = 1₂
Substituting the value of I₂ from Eq. (ii).
I₁ = I₃ cos²\((\theta-\beta)\)
Substituting the value of I₃ from Eq. (i),
I₁ = I₁ cos²13 cos2(\(\theta-\beta\))
or cos²\(\beta \) = 1 [From Eq. (iii), \(\theta=\beta\) ]
\(\beta=0°\) or \(\pi.\)

When a ray after passing through a prism suffers minimum deviation, the ray will travel parallel to the base of the prism inside the prism.

Let PQ and RS are incident and emergent rays. Let incident ray get deviated by \(\delta \) by prism 
\(\angle TMS=\delta \)
Suppose \(\delta _{ 1 },\delta _{ 2 }\) are deviation produced at refractors taking place at AB and AC, respectively.
\(\therefore \delta =\delta _{ 1 }+\delta _{ 2 }\)
\(\delta =\left( i_{ 1 }-r_{ 1 } \right) +\left( i_{ 2 }-r_{ 2 } \right) \)
\(\delta =\left( i_{ 1 }+i_{ 2 } \right) -\left( r_{ 1 }+r_{ 2 } \right) \)
Also, in quadrilateral AQNR,
\(A+\angle QNR=180^{ 0 }\)
[.: QN and RN are normal on two surfaces]
Also in \(\triangle QNR\)
\(\angle QNR+r_{ 1 }+r_{ 2 }=180^{ 0 }\)
\(\Rightarrow A=r_{ 1 }+r_{ 2 }\)
From Eqs. (i) and (ii), we get
\(\delta =\left( i_{ 1 }+i_{ 2 } \right) -A\)
Angle of deviation produced by prism varies with angle of incidence. When prism is adjusted at angle of minimum deviation, then
\(i_{ 1 }=i_{ 2 }=i\)
\(\delta =\delta _{ m }\)
\(\Rightarrow r_{ 1 }=r_{ 2 }=r\)
From Eqs. (i) and (ii), we have
\(\delta _{ m }=2i-2r\quad 2r=A\)
\(\Rightarrow i=\frac { A+\delta _{ m } }{ 2 } \)
\(r=\frac { A }{ 2 } \)
Refractive index of material of prism is
\(\mu =\frac { sini }{ sinr } =\frac { sin\left( \frac { A+\delta _{ m } }{ 2 } \right) }{ sin\frac { A }{ 2 } } \)
This is the required expression.

In Newton's corpuscular (Particle) picture of refraction, particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface. This results in an increase in the normal component of velocity but the component along the surface remains unchanged. This means
\(c\sin { i=\upsilon \sin { r } } \) 
Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
or   \(\frac { \upsilon }{ c } =\frac { \sin { i } }{ \sin { r } } =\mu .\) 
Since \(\mu >i,\therefore \mu >c.\)
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v.
The wave picture of light is consistent with the experimental results.

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\)