In a Young'S Double Slit Experiment, Two Narrow Vertical Slits Placed 0.800 Mm Apart Are Illuminated by the Same Source of Yellow Light of Wavelength 589 Nm. - Physics

Sum

In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?

Solution

Given

Separation between two narrow slits,

$d = 0 . 8 mm = 0 . 8 \times {10}^{- 3} m$

Wavelength of the yellow light,

$\lambda = 589 nm = 589 \times {10}^{- 9} m$

Distance between screen and slit,

$D = 2 . 0 m$

Separation between the adjacent bright bands = width of one dark fringe

That is,

$\beta = \frac{\lambda D}{d}...........(1)$

$\Rightarrow \beta = \frac{589 \times {10}^{- 9} \times 2}{0 . 8 \times {10}^{- 3}}$

$= 1 . 47 \times {10}^{- 3} m$

$= 1 . 47 mm$

Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 17 Light Waves
Q 8 | Page 381