In Young's double-slit experiment, show that the fringe width $\beta$ for interference fringes is $\beta = \dfrac{\lambda D}{d}$, where the symbols have their usual meanings.

**OR**

Draw a labelled diagram to show the image formation at the least distance of distinct vision due

Question

In Young's double-slit experiment, show that the fringe width $\beta$ for interference fringes is $\beta = \dfrac{\lambda D}{d}$, where the symbols have their usual meanings.

**OR**

Draw a labelled diagram to show the image formation at the least distance of distinct vision due to a compound microscope and hence calculate its magnifying power. (1+2=3)

Accepted Answer

![Figure — Hard draw gate ('Draw a labelled diagram to show image formation... compound microscope') on the OR alternativ](/figures/ray-optics-and-optical-instruments/ray-formation-image-compound-microscope-fig-9-24.svg)

>

> [!TLDR]
> Using the small-angle geometry of two coherent slits and a distant screen, the spacing between successive bright fringes works out to $\beta=\lambda D/d$.
>
> **Set-up:** Two coherent sources $S_1,S_2$ separated by $d$, screen at distance $D\gg d$, monochromatic light of wavelength $\lambda$. Let $P$ be a point on the screen at distance $y$ from the central point $O$ (on the perpendicular bisector of $S_1S_2$).
>
> **Step 1 — Path difference**
>
> Using the geometry (with $D\gg d$, small-angle approximation):
>
> $$\Delta = S_2P - S_1P \approx \frac{yd}{D}$$
>
> **Step 2 — Positions of bright fringes**
>
> Constructive interference (bright fringe) occurs when $\Delta = n\lambda$ ($n=0,\pm1,\pm2,\dots$):
>
> $$\frac{y_nd}{D} = n\lambda \quad\Rightarrow\quad y_n = \frac{n\lambda D}{d}$$
>
> **Step 3 — Fringe width**
>
> The fringe width is the distance between two consecutive bright fringes:
>
> $$\beta = y_{n+1}-y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$$
> …

Concept understanding — Fringe Width in Young's Double Slit Experiment

Fringe width and source separation

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