In a Young’s double slit experiment, the separation between the slits is 0.15 m m . In the experiment, a source of light of wavelength 589 n m is used and the interference pattern is observed on a scr

Question

In a Young’s double slit experiment, the separation between the slits is $0.15 m m$ . In the experiment, a source of light of wavelength $589 n m$ is used and the interference pattern is observed on a screen kept $1.5 m$ away. The separation between the successive bright fringes on the screen is:

Options:

Answer:

C

Solution:

$β = \frac{λ D}{d} = \frac{589 \times 10^{- 9} \times 1.5}{0.15 \times 10^{- 3}} = 5.9 m m$

TestHub · Accepted Answer

β = λ D d = 589 × 10 - 9 × 1.5 0.15 × 10 - 3 = 5.9 m m

Physics - Wave Optics Question with Solution | TestHub

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