(a) (i) Define SI unit of power of a lens. (ii) A plano convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex…

Dioptre is the reciprocal of focal length of lens in metre.

Alternatively:

$$\text { Dioptre }=\frac{1}{\text { focal length }(\mathrm{m})}

$$Alternatively:

One dioptre is the power of a lens of focal length of one metre

$$\begin{aligned} & \frac{1}{f}=(\mu-$\displaystyle 1$)\left(\frac{1}{R_{$\displaystyle 1$}}-\frac{1}{R_{$\displaystyle 2$}}\right) & R_{$\displaystyle 1$}=\infty, \quad R_{$\displaystyle 2$}=-$\displaystyle 25$ \mathrm{~cm}, \mu=$\displaystyle 1$ \cdot $\displaystyle 5$ & \frac{1}{f}=(\mu-$\displaystyle 1$)\left(\frac{1}{\infty}+\frac{1}{25}\right) & \text { or } \frac{1}{f}=$\displaystyle 0$ \cdot $\displaystyle 5$ \times \frac{1}{25} \text { or } f=$\displaystyle 50$ \mathrm{~cm} \end{aligned}

$$(ii) $\displaystyle \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Thus the image will be real and inverted

$$\begin{array}{ll} \text { Calculation of (i) distance of second bright fringe } & $\displaystyle 1$ \frac{1}{2} \text { (ii) least distance } & $\displaystyle 1$ \frac{1}{2} \end{array}

$$

$$\begin{aligned} & \therefore n=$\displaystyle 3.5$ & \mathrm{x}_{\mathrm{n}}=\frac{(2 n+1) \lambda D}{2 d}=\frac{($\displaystyle 2$ \times $\displaystyle 3.5$+$\displaystyle 1$) \times $\displaystyle 600$ \times $\displaystyle 10$^{-$\displaystyle 9$} \times $\displaystyle 1$}{$\displaystyle 2$ \times $\displaystyle 0$ \cdot $\displaystyle 6$ \times $\displaystyle 10$^{-$\displaystyle 3$}} & =$\displaystyle 4$ \mathrm{~mm} \end{aligned}