Question

In a double slit experiment, the distance between slits is $5.0mm$ and the slits are $1.0m$ from the screen. Two interference patterns can be seen on the screen : one due to light of wavelength $480nm$ and the other due to light of wavelength $600nm$ . 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
$=(X{)}_{600}-(X{)}_{480}$
$=(3.6\times 10^{-4}m)-(2.88\times 10^{-4}m)$
$=7.2\times 10^{-5}m$
$=0.072\times 10^{-3}m$
$=0.07mm$