In the figure intensity from each slit S 1- S 2 and S 3 is I 0 at P- Wavelength of light is - - D - d List - 1 gives some values and List - 2 gives corresponding intensity at P- then correct match is begin array - -Vhinetext I - rac d - 2 D - lambda - II - frac d - 2 D - frac Vambda 2 - text III - frac d - 2 D - lambda 4 - text U - text None Whhe lend array A- I - P - II - R - III - S - IV - UB- I - P - II - R - III - P - IV - PC- I - P - II - R - III - T - IV - SD- I - P - II - R - III - P - IV - S

The correct option is D I → P, II → R, III → P, IV → SI → P, II → R, III → P, IV → S Path difference for light from S_2 and S_3 is Δ x_23⇒ Δ x_23=√(D^2+4d^ ...

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Byju's Answer

Standard XII

Physics

Energy Levels

In the figure...

Question

In the figure intensity from each slit $S_{1}$ , $S_{2}$ and $S_{3}$ is $I_{0}$ at P. Wavelength of light is $λ$ . (D >> d)
List - 1 gives some values and List - 2 gives corresponding intensity at P, then correct match is
$\begin{matrix} List-1 & List-2 I & \frac{d^{2}}{D} = λ & P & I_{0} II & \frac{d^{2}}{D} = \frac{λ}{2} & Q & 4 I_{0} III & \frac{d^{2}}{D} = \frac{λ}{4} & R & 5 I_{0} IV & \frac{d^{2}}{D} = 2 λ & S & 9 I_{0} T & 3 I_{0} U & None \end{matrix}$

I \to P, I I \to R, I I I \to S, I V \to U

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I \to P, I I \to R, I I I \to P, I V \to P

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I \to P, I I \to R, I I I \to T, I V \to S

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I \to P, I I \to R, I I I \to P, I V \to S

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Solution

The correct option is D $I \to P, I I \to R, I I I \to P, I V \to S$
$I \to P, I I \to R, I I I \to P, I V \to S$

Path difference for light from $S_{2}$ and $S_{3}$ is $Δ x_{23} \Rightarrow$

$Δ x_{23} = \sqrt{D^{2} + 4 d^{2}} - D = D {{(1 + \frac{4 d^{2}}{D^{2}})}^{\frac{1}{2}} - 1} = \frac{2 d^{2}}{D}$
Path difference for light from $S_{1}$ and $S_{3}$ is $Δ x_{13} \Rightarrow$

$Δ x_{13} = \sqrt{D^{2} + 9 d^{2}} - D = D {{(1 + \frac{9 d^{2}}{D^{2}})}^{\frac{1}{2}} - 1} = \frac{9}{2} \frac{d^{2}}{D}$

Corresponding phase difference is $ϕ = \frac{2 π}{λ} Δ x$

$ϕ_{23} = \frac{2 π}{λ} \frac{2 d^{2}}{D} = \frac{4 π}{λ} (\frac{d^{2}}{D})$

$ϕ_{13} = \frac{2 π}{λ} \times \frac{9 d^{2}}{2 D} = \frac{9 π}{λ} (\frac{d^{2}}{D})$

I) $\begin{matrix} ϕ_{23} = \frac{4 π}{λ} \times λ = 4 π ϕ_{13} = \frac{9 π}{λ} \times λ = 9 π \end{matrix} ⎫ ⎬ ⎭ \Rightarrow I = I_{0} (a s A_{R} = A_{0})$

II) $\begin{matrix} ϕ_{23} = \frac{4 π}{λ} \times \frac{λ}{2} = 2 π ϕ_{13} = \frac{9 π}{λ} \times \frac{λ}{2} = \frac{9 π}{2} = 4 π + \frac{π}{2} \end{matrix} ⎫ ⎬ ⎭ \Rightarrow I = 5 I_{0} (a s A_{R} = \sqrt{5 A_{0}^{2}})$

III) $\begin{matrix} ϕ_{23} = \frac{4 π}{λ} \times \frac{λ}{4} = π ϕ_{13} = \frac{9 π}{λ} \times \frac{λ}{4} = \frac{9 π}{4} \end{matrix} ⎫ ⎬ ⎭ \Rightarrow I = I_{0} (a s A_{R} = A_{0})$

IV) $\begin{matrix} ϕ_{23} = \frac{4 π}{λ} \times 2 λ = 8 π ϕ_{13} = \frac{9 π}{λ} \times 2 λ = 18 π \end{matrix} ⎫ ⎬ ⎭ \Rightarrow I = 9 I_{0} (a s A_{R} = 3 A_{0})$
For all above optioin,
$I_{0} \propto A_{0}^{2}$ , $I \propto A_{R}^{2}$

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Similar questions

In the figure intensity from each slit

S_{1}

S_{2}

and

S_{3}

I_{0}

at P. Wavelength of light is

λ

(D >> d)
List - 1 gives some values and List - 2 gives corresponding intensity at P. If space between slits and screen is filled with fluid of refractive index

\frac{3}{2}

then correct match is

\begin{matrix} List-1 & List-2 I & \frac{d^{2}}{D} = λ & P & I_{0} II & \frac{d^{2}}{D} = \frac{λ}{2} & Q & 4 I_{0} III & \frac{d^{2}}{D} = 1.5 λ & R & 5 I_{0} IV & \frac{d^{2}}{D} = 2 λ & S & 9 I_{0} T & 3 I_{0} U & None \end{matrix}

Q. For a given YDSE experiment, List - 1 gives distance of a point from center

C

of the screen in

m m

and List - 2 gives resultant intensity at that point.

I_{0}

is intensity due to each slit at screen. Slit separation is

d = 0.2 m m

. Distance of screen from slits is

D = 1 m m

and wavelength of light is

λ = 400 n m

List -1	List - 2
(I) $y = 0$	(P) 0
(II) $y = 0.5$	(Q) $I_{0}$
(III) $y = 1$	(R) $2 I_{0}$
(IV) $y = \frac{4}{3}$	(S) $3 I_{0}$
	(T) $4 I_{0}$
	(U) None

If slit

S_{1}

and slit

S_{2}

are covered by thin films each of thickness

2000 n m

and of refractive index 1.5 and 1.2 respectively, then match List - I with List - II

\begin{matrix} Column-I & Column-II i) 0.5 mole of S O_{2} (g) & p) Occupies 11.2 L at 1 atm and 273 K ii) 1 g of H_{2} (g) & q) Weighs 24 g iii) 0.5 mole of O_{3} (g) & r) Total atoms = 1.5 \times N_{A} iv) 1 g molecule of O_{2} (g) & s) Weighs 32 g \end{matrix}

\begin{matrix} List-I & List-II (I) & Work function & (P) & Wavelength of photon used less than threshold wavelength (I I) & Threshold frequency & (Q) & Photo intensity of light (I I I) & Stopping potential & (R) & number of photons (I V) & Photo current & (S) & Metal (T) & Electron (U) & Surface of liquid \end{matrix}

Q. A fast moving neutron collides with a Lithium atom at rest in ground state with kinetic energy K. List - 1 gives different values of K and List- 2 gives various possible phenomena

\begin{matrix} List-1 & List-2 I & K=81.6 eV & P & Elastic collision may take place II & K=104.9 eV & Q & Elastic collision will take place III & K=108.8 eV & R & Inelastic collision may take place IV & K=200 eV & S & Perfectly inelastic collision may take place T & Atom may get ionised U & Photon may be emitted \end{matrix}

Match entries in List - 1 with those in List - 2.

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In the figure intensity from each slit S1, S2 and S3 is I0 at P. Wavelength of light is λ. (D >> d) List - 1 gives some values and List - 2 gives corresponding intensity at P, then correct match is List-1List-2Id2D=λPI0IId2D=λ2Q4I0IIId2D=λ4R5I0IVd2D=2λS9I0T3I0UNone