Question 245: Wave Optics

Question

In Young's double slit experiment the wavelength of light used is $6000 \mathop {\rm{A}}\limits^{\rm{o}}$, the screen is 40 cm from the slits and the fringe width is 0.012 cm , the distance between two slits is

Accepted Answer

Given:
Wavelength of light:
$\lambda = 6000 \,	ext{ÃÂ} = 6000 	imes 10^{-10}\,	ext{m} = 6 	imes 10^{-7}\,	ext{m}$
Distance between slits and screen:
$D = 40 \,	ext{cm} = 0.40 \,	ext{m}$
Fringe width:
$\beta = 0.012 \,	ext{cm} = 1.2 	imes 10^{-4}\,	ext{m}$
We need the distance between the two slits: $d$.
Formula:
$ \beta = \frac{\lambda D}{d} \quad \Rightarrow \quad d = \frac{\lambda D}{\beta} $
Substitution:
$ d = \frac{(6 	imes 10^{-7})(0.40)}{1.2 	imes 10^{-4}} $
$ d = \frac{2.4 	imes 10^{-7}}{1.2 	imes 10^{-4}} $
$ d = 2.0 	imes 10^{-3}\,	ext{m} $
Convert to cm:
$ 2.0 	imes 10^{-3}\,	ext{m} = 0.20 \,	ext{cm} $
Final Answer:
Option D: 0.2 cm
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Answer

0.024 cm

Answer

2.4 cm

Answer

0.24 cm

Answer

0.2 cm

Wave Optics

In Young's double slit experiment the wavelength of light used is 6000 \mathop {\rm{A}}\limits^{\rm{o}}, the screen is 40 cm from the slits and the fringe width is 0.012 cm , the distance between two slits is

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