Question

Light of $630\mathrm{nm}$wavelength illuminates two slits that are$0.25\mathrm{mm}$apart. FIGURE $\mathrm{EX}33.5$shows the intensity pattern seen on a screen behind the slits. What is the distance to the screen$?$

Short answer

The distance to the screen is$1.31\mathrm{m}$.

Step-by-step solution

Formula for spacing between the peaks

In the double slit experiment, the position of the brilliant fringes can be written as follows:

${\mathrm{y}}_{\mathrm{m}}=\frac{\mathrm{m\lambda L}}{\mathrm{d}}$

As a result, the distance between the peaks of any two brilliant fringes is

localid="1649184414720" $\mathrm{\Delta y}={\mathrm{y}}_{\mathrm{m}+1}-{\mathrm{y}}_{\mathrm{m}}$

$=\frac{(\mathrm{m}+1)\mathrm{\lambda L}}{\mathrm{d}}-\frac{\mathrm{m\lambda L}}{\mathrm{d}}$

$=\frac{\mathrm{\lambda L}}{\mathrm{d}}$

Calculation of distance to the screen

To isolate$\mathrm{L}$, rearranging the equation yields,

$\mathrm{L}=\frac{\mathrm{d\Delta y}}{\mathrm{\lambda }}$

localid="1649184047902" $∆\mathrm{y}$is the distance between any two succeeding peaks in the given figure, which islocalid="1649184044408" $0.33\mathrm{m}$.

Thus

$\mathrm{L}=\frac{\left(0.25\times {10}^{-3}\mathrm{m}\right)\times 0.33\times {10}^{-2}\mathrm{m}}{630\times {10}^{-9}\mathrm{m}}$

$\mathrm{L}=1.31\mathrm{m}$