Question

In YDSE, light of wavelength 60 nm is used. The separation between the sources is 6 mm and between the sources and the screen is 2 m. Find the position of a point lying between third maxima and third minima where the intensity is three-fourth of the maximum intensity on the screen.

Answer 1

Answer:

the position of a point lying between third maxima and third minima where the intensity is three-fourth of the maximum intensity on the screen = 3.3 micrometer.

Explanation:

Given:

Distance between the sources and the screen is = $D=2m$

The separation between the sources is = $d = 6mm$

Formula of intensity for YDSE is given by,

$I=4I_ocos^2(\frac{\phi}{2} )$ -----(1)

Where, $\phi$ denotes phase

and $I$ is the phase difference.

Maximum intensity = $4I_o$      [at $\phi = 0^o$]

When intensity is three-fourth of maximum intensity = $I=\frac{3}{4}*4I_o=3I_o$

Equating it with equation (1):

$3I_o=4I_ocos^2(\frac{\phi}{2} )$

$\implies cos^2(\frac{\phi}{2} )=\frac{3}{4}$

$\implies \phi = 60^o = \frac{\pi}{3} radians$

Also, we know,

phase difference ($\phi$) = $\frac{2\pi}{\lambda} *$ path difference ($\Delta x$)

$\implies$ path difference ($\Delta x$) = $\frac{\lambda}{2\pi}*\phi$

path difference = $\frac{\lambda}{2\pi}*\frac{\pi}{3}$ = $\frac{\pi}{6}$

Position of the point on the screen is given by the formula:

$y=\frac{\Delta x D}{d}$

$y=\frac{\pi}{6} *\frac{2}{60*10^{-9}}$

$\implies y=3.3 *10^{-6}m$

#SPJ2

Answer 2

$3.3$ micrometer is the location of the point between the third maxima and third minima where the intensity is three-fourths of the highest intensity on the screen.

Explanation:

Given - Distance between the sources and the screen is $= D = 2m$.

The separation between the sources is $= d = 6mm$.

The formula of intensity for YDSE -

$I=4I_{0} cos^{2} ($ Ф$/2)$     ...... (1)

Ф denotes phase and $I$ is the phase difference.

Maximum intensity $= 4I_{0}$.

When intensity is three-fourths of maximum intensity $= I = \frac{3}{4 }$ × $4I_{0} = 3I_{0}$.

Equation this with equation (1) as follows:

$3I_{0} = 4I_{0} cos^{2} ($Ф$/2)$

⇒ $cos^{2} ($Ф$/2) = \frac{3}{4}$

⇒ Ф $= 60^{0} = \frac{\pi }{3} radians$

We also know that phase difference (Ф) $=2\pi /$λ path difference (Δ$x$).

⇒ Path difference (Δ$x$) = λ$/2\pi$ × Ф

Path difference = λ$/2\pi$ ×$\frac{\pi }{3} = \frac{\pi }{6}$.

The following formula determines where the point is located on the screen:

$y =$ Δ${xd}/{d}$

$y = \frac{\pi }{6}$ × $\frac{2}{60*10x^{-9} }$

⇒ $y = 3.3$ × $10^{-6} m$

Therefore, $3.3$ micrometer is the location of the point between the third maxima and third minima where the intensity is three-fourths of the highest intensity on the screen.

#SPJ2