In YDSE apparatus as shown in figure wavelength of light used in λ . The screen is moved away from the source with a constant speed V. Initial distance between screen and plane of slits was D . Suppose P is the point where 5t h order maxima was lying at t=0, then after how much time third-order maxima will lie at this point P on that screen?

Question

In YDSE apparatus as shown in figure wavelength of light used in λ . The screen is moved away from the source with a constant speed V. Initial distance between screen and plane of slits was D . Suppose P is the point where 5t h order maxima was lying at t=0, then after how much time third-order maxima will lie at this point P on that screen? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-dkadkdiwaefdhklj.jpg" /> (A) (2 D/V) (B) (2 D/3 V) (C) (3 D/2 V) (D) (3 D/V)

Tardigrade Admin · Accepted Answer

At t=0, Δ x=5λ =(y d/D) .....(i) After t times Δ x1=3λ =(y d/D + v t) ......(ii) Dividing (i) and (ii) we get (5/3)=(D + v t/D) 5D=3D+3vt 2D=3vt ⇒ t=(2 D/3 v)

$\frac{2 D}{V}$

$\frac{2 D}{3 V}$

$\frac{3 D}{2 V}$

$\frac{3 D}{V}$

At $t = 0, Δ x = 5 λ = \frac{y d}{D}$ .....(i)
After t times $Δ x_{1} = 3 λ = \frac{y d}{D + v t}$ ......(ii)
Dividing (i) and (ii) we get
$\frac{5}{3} = \frac{D + v t}{D}$
$5 D = 3 D + 3 v t$
$2 D = 3 v t \Rightarrow t = \frac{2 D}{3 v}$

V2D​

3V2D​

2V3D​

V3D​

At t=0,Δx=5λ=Dyd​ .....(i) After t times Δx1​=3λ=D+vtyd​ ......(ii) Dividing (i) and (ii) we get 35​=DD+vt​ 5D=3D+3vt 2D=3vt⇒t=3v2D​

$\frac{2 D}{V}$

$\frac{2 D}{3 V}$

$\frac{3 D}{2 V}$

$\frac{3 D}{V}$

At $t = 0, Δ x = 5 λ = \frac{y d}{D}$ .....(i)
After t times $Δ x_{1} = 3 λ = \frac{y d}{D + v t}$ ......(ii)
Dividing (i) and (ii) we get
$\frac{5}{3} = \frac{D + v t}{D}$
$5 D = 3 D + 3 v t$
$2 D = 3 v t \Rightarrow t = \frac{2 D}{3 v}$