(a) (i) Draw a ray diagram for the formation of the image of an object by a convex mirror. Hence, obtain the mirror equation. (ii) Why are…

For paraxial rays MP can be considered to be a straight line perpendicular to CP. Therefore right angled triangles $\displaystyle A^{\prime} B^{\prime} F$ and MPF are similar:

$\displaystyle \frac{B^{\prime} A^{\prime}}{P M}=\frac{B^{\prime} F}{F P}$

Or $\displaystyle \frac{B^{\prime} A^{\prime}}{B A}=\frac{B^{\prime} F}{F P}$ ($\displaystyle \because P M=A B$) ----------------($\displaystyle 1$)

Since $\displaystyle \angle A P B=\angle A^{\prime} P B^{\prime}$, the right angled triangles $\displaystyle A^{\prime} P B^{\prime}$ and ABP are also similar:

Therefore $\displaystyle \frac{B^{\prime} A^{\prime}}{B A}=\frac{B^{\prime} P}{B P}$ ----------------------------------- ($\displaystyle 2$)

Comparing eq ($\displaystyle 1$) and ($\displaystyle 2$), we get:

$\displaystyle \frac{B^{\prime} F}{F P}=\frac{B^{\prime} P}{B P}$

$\displaystyle \frac{P F-P B^{\prime}}{F P}=\frac{B^{\prime} P}{B P}$

Using sign convention: $\displaystyle P F=f$, $\displaystyle P B^{\prime}=+v$, $\displaystyle P B=-u$

On solving: $\displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

To improve image quality by minimizing various optical aberrations in lenses.

Magnification produced by compound microscope:

$\displaystyle m=m_{0} \times m_{e}$

$\displaystyle m_{0}=\frac{m}{m_{e}}=\frac{m}{\dfrac{D}{f_{e}}}$

$\displaystyle m_{0}=\frac{200}{\dfrac{25}{2}}=16$

Wavefront is a surface of constant phase.

OR: Locus of points which oscillate in phase.

Ray – the straight line path along which light travels (or energy propagates).

OR: Ray is normal to the wavefront.

Huygens' Principle: Each point of the wavefront is the source of secondary disturbance and the wavelets emanating from the points spread out in all directions with the speed of the wave. The wavelets emanating from the wavefront are usually referred to as secondary wavelets. A common tangent to all these spheres gives the new position of the wavefront at a later time.

Triangles EAC and BAC are congruent, therefore $\displaystyle \angle i=\angle r$.

Position of 4th bright fringe:

$\displaystyle x_{4(\text {bright})}=\frac{4 D \lambda}{d}$

Position of 2nd dark fringe:

$\displaystyle x_{2(\text {dark})}=\frac{3}{2} \frac{D \lambda}{d}$

$\displaystyle x_{4(\text {bright})}-x_{2(\text {dark})}=5 \mathrm{~mm}$

$\displaystyle \frac{4 D \lambda}{d}-\frac{3}{2} \frac{D \lambda}{d}=5 \times 10^{-3}$

$\displaystyle \lambda=6 \times 10^{-6} \mathrm{~m}$