In a double-slit experiment, the fifth maximum is 2.8 cm from the central maximum on a screen that is 1.5 m away from the slits. If the slits are 0.15 mm apart, what is the wavelength of the light being used?

Answer :

Answer:

Explanation:

Screen distance D = 1.5 m

slit separation d = .15 x 10⁻³ m

distance of 5 th maxima   = 2.8 x 10⁻² m

distance of n th maxima = n x λ D/d

2.8 x 10⁻² = 5 x ( λ 1.5 / .15 x 10⁻³ )

λ  = (2.8 x 10⁻² x .15 x 10⁻³ ) / (5 x 1.5 )

= .42 x 10⁻⁵ / 7.5

.056 x 10⁻⁵

= 56 x 10⁻⁸ m

= 5600 A

Answer:

The wavelength of the light is 560 nm.

Explanation:

Given that,

Distance from the central maximum y= 2.8 cm

Distance D=1.5 m

Distance between the slit = 0.15 m

We need to calculate the wavelength of the light

Using formula of constructive interference

[tex]m\lambda=\dfrac{yd}{D}[/tex]

[tex]\lambda=\dfrac{yd}{mD}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{2.8\times\times10^{-2}\times0.15\times10^{-3}}{5\times1.5}[/tex]

[tex]\lambda=5.6\times10^{-7}[/tex]

[tex]\lambda=560\ nm[/tex]

Hence, The wavelength of the light is 560 nm.

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