Question

Two narrow parallel slits separated by $0.850mm$ are illuminated by $600nm$ light, and the viewing screen is $2.80m$ away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point $2.50mm$ from the central bright fringe? (b) What is the ratio of the intensity at this point to the intensity at the centre of a bright fringe?

Answer 1

Answer:

(a) $7.95radian$, (b) $0.453$.

Explanation:

y = Distance from central bright fringe = 2.5 mm

λ = Wavelength = 600 nm

L = Distance between screen and source = 2.8 m

d = Slit distance = 0.85 mm

$tan\theta =y/L$

$\Rightarrow tan\theta =2.5/2800=0.000892$

$\Rightarrow \theta =ta{n}^{-1}0.000892={0.05115}^{\circ }$

$\Delta r=dsin\theta =7.589\times {10}^{-7}$

a) Phase difference

$\varphi =\frac{2\pi }{\lambda }\Delta r$

$\Rightarrow \varphi =\frac{2\pi }{6\times {10}^{-9}}\times 7.589\times {10}^{-7}=7.947rad$

∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians

b) $\frac{I}{I_{max}}=co{s}^2\frac{\varphi }{2}$

$\Rightarrow \frac{I}{I_{max}}=co{s}^2\frac{7.947}{2}$

$\Rightarrow \frac{I}{I_{max}}=0.4535$

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe 0.4535