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(D) The formula for fringe width in $YDSE$ is given by $\omega = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between th

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(D) The formula for fringe width in $YDSE$ is given by $\omega = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the separation between the slits.Given that the new separation $d' = \frac{d}{2}$ and the new distance $D' = 2D$.The new fringe width $\omega'$ will be $\omega' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(d/2)} = 4 	imes \frac{\lambda D}{d} = 4\omega$.Therefore,the fringe width becomes $4$ times the original value.

In $YDSE$,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

Similar Questions

Two slits are made $1\, mm$ apart and the screen is placed $1\, m$ away. The fringe separation when monochromatic light of wavelength $500\, nm$ is used is

Which of the following statements is incorrect regarding interference fringes?

In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

In Young's double slit experiment,the distance between the slits is $1 \,mm$ and the distance between the slit and the screen is $1 \,m$. If the $10^{th}$ fringe is $5 \,mm$ away from the central bright fringe,then the wavelength of the light used will be.....$\mathring A$.

Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other.
Reason : The fringe width is inversely proportional to the distance between the two slits.

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