(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)

 

(a) The speed of light in vacuum is a universal constant, independent of all the factors listed and anything else.
(b) Dependence of the speed of light in a medium:
(i) does not depend on the nature of the source 
(wave speed is determined by the properties of the medium of propagation.This is also true for other waves e.g. sound waves etc.)
(ii) independent of the direction propagation for isotropic media.
(iii) Independent of the motion of the source relative to the medium.
(iv) depends on wavelength.
(v) independent of intensity.

The total apparent shift in the position of the image due to all the three media is
given by
d = t₁[1-1/(\({ \mu }\)₁) + t₂[1-1/(\({ \mu }\)₂) + t₃[1-1/(\({ \mu }\)₃)
Given t₁ = 4.0 cm, t₂ = 6.0 cm , t₃ = 8.0 cm
\({ \mu }\)1 = 1.5 , \({ \mu }\)2=1.4 , \({ \mu }\)3 = 1.3 cm
d = 4.0(1-1/1.5) + 6.0(1-1/1.4) + 8.0(1-1/1.3)
= 1.33 + 1.71 + 1.85 = 4.89 cm

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  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Here,
\(u_{ 1 }=-60\ cm,\ m_{ 1 }=\frac { 1 }{ 2 }\)
\( u_{ 2 }=?,\ m_{ 2 }=\frac { 1 }{ 3 } \)
As \(m_{ 1 }=-\frac { \upsilon _{ 1 } }{ u_{ 1 } } \therefore \frac { 1 }{ 2 } =-\frac { \upsilon _{ 1 } }{ -(-60) } ,\upsilon _{ 1 }=30\quad cm\)
From \(\frac { 1 }{ f } =\frac { 1 }{ \upsilon _{ 1 } } +\frac { 1 }{ u_{ 1 } } =\frac { 1 }{ 30 } -\frac { 1 }{ 60 } =\frac { 1 }{ 60 } \)
\(f=60\ cm\)
Again, \(m_{ 2 }=-\frac { \upsilon _{ 2 } }{ u_{ 2 } } =\frac { 1 }{ 3 } ,\ \upsilon _{ 2 }=-\frac { u_{ 2 } }{ 3 } \)
From
\(\frac { 1 }{ \upsilon _{ 2 } } +\frac { 1 }{ u_{ 2 } } =\frac { 1 }{ f }\)
\( \frac { 1 }{ u_{ 2 } } =\frac { 1 }{ f } -\frac { 1 }{ \upsilon _{ 2 } } =\frac { 1 }{ 60 } +\frac { 3 }{ u_{ 2 } } \)
or \(\frac { 1 }{ u_{ 2 } } -\frac { 3 }{ u_{ 2 } } =\frac { 1 }{ 60 } ,\ u_{ 2 }=-120 \ cm\)​​​​​​

Let \(\mu \) be refractive index of the transparent surface. If \({ i }_{ p }\)is polarising angle r is the angle of refraction as shown in Fig. then from Snell law, we have 
\( \mu =\frac { \sin { { i }_{ p } } }{ \sin { { r } } } ....(1)\)
\(From \ Breester' \ law, \ we \ have\)
\(\mu =\tan { { i }_{ p } } =\frac { \sin { { i }_{ p } } }{ \sin { { i }_{ p } } } ....(2)\)
\(From \ Eqs.\ (1)\ (2), \ we \ have\)
\(\frac { \sin { { i }_{ p } } }{ \sin { r } } =\frac { \sin { { i }_{ p } } }{ { \cos { { i }_{ p } } } }\)
\(or \ \sin { r } =\cos { { i }_{ p } } \)
\(or \ \sin { r } =\sin { ({ 90 }^{ \circ } } -{ i }_{ p })\)
\(or \ r ={ 90 }^{ \circ }-{ i }_{ p }\)
\( or \ { i }_{ p }+r={ 90 }^{ \circ }\)
i.e. the reflected and refracted beams of light at polarising angle are perpendicular to each other.

A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

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.

Real depth = y cm
Apparent depth = y- x cm
Refractive index of oil,
\({ \mu }\) = real depth/ apparent depth = y / y-x

The refractive index of diammond is much higher than that of  glass. Due to high refractive index, the critical angle for diamond-air interface is low. The diamond is cut suitably, so that the light entering the diamond from any face suffers multiple total internal reflections at the various surfaces. This gives sparklingeffect to the diamonds.

Here,   
\(\lambda =6563\mathring { A } ,\Delta \lambda =+15\mathring { A }\)
\( \\ C=3\times { 10 }^{ 8 }{ ms }^{ -1 }\)
 Since, the star is receding away,hence its velocity v is negative.
\(\Delta \lambda =\frac { v\lambda }{ c } or\quad v=-\frac { c\Delta \lambda }{ \lambda } \)
\(\\ =-\frac { 3+{ 10 }^{ 8 }\times 15 }{ 6563 } =-6.86\times { 10 }^{ 5 }{ 10 }ms^{ -1 }\)
Negative sign shows that star is receding away from the earth.

\(3800\overset { \circ }{ A } \ to \ 7600\overset { \circ }{ A } \)

Light waves reflected from outer and inner surface of soap bubble interfere. For different wavelengths, conditions for constructive interference are satisfied at different positions. That is why beautiful colours are seen.

A myopic person has to use spectacles with concave lens.He may have normal near point (\(\approx\)25cm). To read with specs, he has o hold the book at a distance greater than 25cm. As angular size of object at a distance > 25 cm is less than angular size of object at 25 cm, therefore, the person prefers to remove his spectacles while reading.

The focal length of mirror does not change on changing the wavelength of light incident on it. This is because f = R/2, i.e. f depends only on radius of curvature of the mirror.

No, because the image formed by a convex mirror is virtual.

Visual angle is the angle subtended by an object at the eye.