Fluorescence is the phenomenon of emission of energy in visible region by a substance on absorbing radiations of higher frequency.If the process of emission continues even after the exciting radiations are stopped, the phenomenon is called phosphorescence.

\(m=1+{d\over f}\).As f_r>f_v, therefore, magnifying power increases when violet light is used. 

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.

This problem is due to astigmatism of the eye. This defect is removed using a cylindrical lens. With appropriate axis and suitable radius of curvature.

Range of the vision is the distance between near point and far point of eye.

1.33; 1.7 cm

f_e = 2 cm, f_o = 8 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.