In a Young's double slit experiment the distance between slits is increased five times where as their distance from screen in halved, then the fringe width is

Question

In a Young's double slit experiment the distance between slits is increased five times where as their distance from screen in halved, then the fringe width is (A) becomes (1/90) (B) becomes (1/20) (C) becomes (1/10) (D) it remains same

Tardigrade Admin · Accepted Answer

Fringe width, β=(λ D/d) ...(i) According to the question, D prime=(D/2) text and d prime=5 d β prime=(D prime λ/d prime)=(((D/2)) λ/5 d)=(1/10) (D λ/d) β prime=(β/10) [from Eq. (i)]

Q. In a Young's double slit experiment the distance between slits is increased five times where as their distance from screen in halved, then the fringe width is

becomes $\frac{1}{90}$

becomes $\frac{1}{20}$

becomes $\frac{1}{10}$

it remains same

Fringe width, $β = \frac{λ D}{d}$ ...(i)
According to the question,
$D^{'} = \frac{D}{2} and d^{'} = 5 d$
$β^{'} = \frac{D ^{'} λ}{d ^{'}} = \frac{( \frac{D}{2} ) λ}{5 d} = \frac{1}{10} \frac{D λ}{d}$
$β^{'} = \frac{β}{10}$ [from Eq. (i)]

Q. In a Young's double slit experiment the distance between slits is increased five times where as their distance from screen in halved, then the fringe width is

becomes 901​

becomes 201​

becomes 101​

it remains same

Fringe width, β=dλD​ ...(i) According to the question, D′=2D​ and d′=5d β′=d′D′λ​=5d(2D​)λ​=101​dDλ​ β′=10β​ [from Eq. (i)]

becomes $\frac{1}{90}$

becomes $\frac{1}{20}$

becomes $\frac{1}{10}$

Fringe width, $β = \frac{λ D}{d}$ ...(i)
According to the question,
$D^{'} = \frac{D}{2} and d^{'} = 5 d$
$β^{'} = \frac{D ^{'} λ}{d ^{'}} = \frac{( \frac{D}{2} ) λ}{5 d} = \frac{1}{10} \frac{D λ}{d}$
$β^{'} = \frac{β}{10}$ [from Eq. (i)]