Question

In a double slit experiment, the distance between slits is $ 5.0 \,mm $ and the slits are $ 1.0\, m $ from the screen. Two interference patterns can be seen on the screen : one due to light of wavelength $ 480 \,nm $ and the other due to light of wavelength $ 600 \,nm $ . What is the separation on the screen between the third order bright fringes of the two interference patterns ?

• A. 0.02 mm
• B. 0.05 mm
• C. 0.07 mm
• D. 0.09 mm

Answer 1

Answer: (C)

Position of bright fringes is given by $\lambda = n\lambda \frac{D}{d}$
For $3$rd order bright fringes, $n = 3$
According to question,
$(X)_{480} = \frac{3\times (480\times 10^{-9}) \times 1}{(5\times 10^{-3})} $
$ = 2.88 \times 10^{-4}\,m$
$(X)_{600} = \frac{3\times (600 \times 10^{-9}) \times 1}{(5\times 10^{-3})}$
$= 3.6 \times 10^{-4}\,m$
$\therefore $ Separation between third order bright is
$ = (X)_{600} - (X)_{480}$
$= (3.6 \times 10^{-4}\,m) - ( 2.88 \times 10^{-4} \,m)$
$ = 7.2 \times 10^{-5} \,m$
$ = 0.072 \times 10^{-3}\,m$
$= 0.07\,mm$