According to Brewster's law,
\(\mu =tanp....(i)\)
\(Also \ \mu =\frac { 1 }{ sin\quad C } .....(ii)\)
\( From \ (i) \ and \ (ii), \ tanp=\frac { 1 }{ sin \ C } \)
\(If \ r \ is \ angle \ of \ refraction, \ then \ p=90°-r\)
\(or \ tan \ r=sin \ C \ r={ tan }^{ -1 }(sin \ C)\)

Here R = 2m,
\(f=\frac { R }{ 2 } =\frac { 2 }{ 2 } =1m\)
Using morror formula, we have
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \Rightarrow \frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u }\)
\( \Rightarrow \frac { 1 }{ v } =\frac { u-f }{ fu } \Rightarrow v=\frac { fu }{ u-f } \)
When Jogger is 39 m away, then u = -39m
\(\Rightarrow \) we get
\(v=\frac { fu }{ u-f } =\frac { 1\left( -39 \right) }{ -39-1 } or \ v=\frac { 39 }{ 40 } m\)
As, the Jogger is running at a constant speed of 5 m/s, after 1 s, the positionof the image(v) for
u = -39 + 5
u = -34m
Again using the Equation we get
\(\Rightarrow v=\frac { fu }{ u-f } =\frac { 1\left( -34 \right) }{ -34-1 } or\quad v=\frac { 34 }{ 35 } m\)
Difference in apparent position of Jogger in 1s
\(=\frac { 39 }{ 40 } -\frac { 34 }{ 35 } =\frac { 1365-1360 }{ 1400 } =\frac { 1 }{ 280 } m\)
Average speed of Joggers image \(=\frac { 1 }{ 280 } m/s\)
Similarly, for u = -29m, -19m and -9 m, average speed of Jogger image is \(\frac { 1 }{ 150 } m/s\),\(\frac { 1 }{ 60 } m/s\),\(=\frac { 1 }{ 10 } m/s\) respectively
The speed increases as the Jogger approaches the car>
This can be experienced by the person in the car.

(i) The values shown by him are as follows:
(a) Consulting others in case of need
(b) Curosity
(c) Sharing knowledge.
(ii) From these three lenses, he will use a lens of power 0.5D for objective and lens of power 10D for eyepiece.

(ii) As, the number of point sources increases, their contribution towards intensity also increases. Intensity varies as square of the slit width. Thus, when the width of the slit is made double the original width, intensity will get four times of its original value.
Width of central maximum is given by \(\beta =\frac { 2D\lambda }{ b } \)
Where, D = distance between screen and slit,
 \(\lambda \) = wavelength of the light,b = size of slit.

So, with the increase in size of slit, the width of central maxima decreases. Hence, double the size of the slit would result as half the width of the central maxima.

(i) Here,
\(\lambda =600 \ nm=600\times { 10 }^{ -9 }m=6\times { 10 }^{ -7 }m\)
\(\theta ={ 0.1 }^{ \circ }=\frac { 0.1\pi }{ 180 } rad,d=?\)
From angular width,\(\theta =\frac { \lambda }{ d } \)
\(\Rightarrow \ d=\frac { \lambda }{ \theta } =\frac { 6\times { 10 }^{ -7 } }{ \frac { \pi }{ 180 } \times 0.1 } =3.44\times { 10 }^{ -4 }m\)
(ii) The frequency and wavelength of reflected wave will not change. The refracted wave will have same frequency. The velocity of light in water is given by \(v=f\lambda \)
where, v = velocity of light
f = frequency of light
​​​​​​​​​​​​\( ​​lambda =wavelength \ of \ light\)
If velocity will decrease, then wavelength (\(\lambda \)) will also decrease.