A light source, which emits two wavelengths λ1=400 nm and λ2=600 nm, is used in a Young's double slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then

Question

A light source, which emits two wavelengths λ1=400 nm and λ2=600 nm, is used in a Young's double slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then (A) β2 > β1 (B) m1 > m2 (C) From the central maximum, 3 text rd maximum of λ2 overlaps with 5 text th minimum of λ1 (D) the angular separation of fringes of λ1 is greater than λ2

Tardigrade Admin · Accepted Answer

β=( D λ/ d ) ∵ λ2>λ1 ⇒ β2>β1 Also m 1 β1= m 2 β2 ⇒ m 1> m 2 Also 3(( D / d ))(600 nm ) =(2 × 5-1)(( D /2 d )) 400 nm Angular width θ=(λ/ d )

$β_{2} > β_{1}$

$m_{1} > m_{2}$

From the central maximum, $3^{rd}$ maximum of $λ_{2}$ overlaps with $5^{th}$ minimum of $λ_{1}$

the angular separation of fringes of $λ_{1}$ is greater than $λ_{2}$

$β = \frac{D λ}{d}$
$∵ λ_{2} > λ_{1}$
$\Rightarrow β_{2} > β_{1}$
Also $m_{1} β_{1} = m_{2} β_{2}$
$\Rightarrow m_{1} > m_{2}$
Also $3 (\frac{D}{d}) (600 nm)$
$= (2 \times 5 - 1) (\frac{D}{2 d}) 400 nm$
Angular width $θ = \frac{λ}{d}$

β2​>β1​

m1​>m2​

From the central maximum, 3rd maximum of λ2​ overlaps with 5th minimum of λ1​

the angular separation of fringes of λ1​ is greater than λ2​

β=dDλ​ ∵λ2​>λ1​ ⇒β2​>β1​ Also m1​β1​=m2​β2​ ⇒m1​>m2​ Also 3(dD​)(600nm) =(2×5−1)(2dD​)400nm Angular width θ=dλ​

$β_{2} > β_{1}$

$m_{1} > m_{2}$

From the central maximum, $3^{rd}$ maximum of $λ_{2}$ overlaps with $5^{th}$ minimum of $λ_{1}$

the angular separation of fringes of $λ_{1}$ is greater than $λ_{2}$

$β = \frac{D λ}{d}$
$∵ λ_{2} > λ_{1}$
$\Rightarrow β_{2} > β_{1}$
Also $m_{1} β_{1} = m_{2} β_{2}$
$\Rightarrow m_{1} > m_{2}$
Also $3 (\frac{D}{d}) (600 nm)$
$= (2 \times 5 - 1) (\frac{D}{2 d}) 400 nm$
Angular width $θ = \frac{λ}{d}$