Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The location of the central maximum is marked on the screen and labeled y = 0.

A) At what distance, on either side of y = 0, are the m = 1 bright fringes ?

B) A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by 5.0

B) A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by 5.0 * 10⁻¹⁶ s in comparison to the wave going through the other slit. What fraction of the period of the light wave is this delay?

Answer:

A) y = 3.0 mm

B) Fraction of delay = 0.25

Explanation:

wavelength of the light, [tex]\lambda = 600 nm = 600 * 10^{-9} m[/tex]

Separation between the two slits, d = 0.22 mm = 0.00022 m

Distance between the screen and the slit, L = 1.1 m

The distance of the bright fringes, [tex]y = \frac{m \lambda L}{d}[/tex]

[tex]y = \frac{1 * 600 * 10^{-9}* 1.1 }{0.00022}[/tex]

[tex]y = 0.003 m[/tex]

[tex]y = 3.0 mm[/tex]

b) The period of the light wave in air can be calculated by using the formula

[tex]c = \frac{\lambda}{T} \\T = \frac{\lambda}{c}[/tex]

c = 3 * 10⁸ m/s

[tex]T = \frac{600 * 10^{-9} }{3 * 10^{8} }[/tex]

[tex]T = 2 * 10^{-15} sec[/tex]

The time delay due to the slow travel of light in glass, [tex]\triangle t = 5 * 10^{-16} s[/tex]

Fraction of delay = [tex]\frac{5 * 10^{-16} }{2 * 10^{-15} }[/tex]

Fraction of delay = 0.25

Cricetus Cricetus · Answer 2 · 2021-12-08T08:19:13+01:00

(a) The distance will be "3.0 mm".

(b) The required fraction will be "0.25".

Given:

Wavelength, [tex]\lambda = 600 \ nm[/tex] or, [tex]600\times 10^{-9} \ m[/tex]
Length, [tex]L = 1.1 \ m[/tex]
Distance, [tex]d = 0.22 \ mm[/tex] or, [tex]0.22\times 10^{-3} \ m[/tex]

(a) The position of 1st order maxima will be:

→ [tex]y = \frac{\lambda L}{d}[/tex]

By substituting the values, we get

[tex]= \frac{(600\times 10^{-9})(1.1)}{0.22\times 10^{-3}}[/tex]

[tex]= 3\times 10^{-3}[/tex]

[tex]= 3.0 \ mm[/tex]

(b) The light wave's period will be:

→ [tex]T = \frac{\lambda}{c}[/tex]

By putting the values, we get

[tex]= \frac{600\times 10^{-9}}{3.0\times 10^8}[/tex]

[tex]= 2.0\times 10^{-15} \ s[/tex]

hence,

The required fraction will be:

→ [tex]f = \frac{\Delta t}{T}[/tex]

By substituting the values, we get

[tex]= \frac{5.0\times 10^{-16}}{2.0\times 10^{-15}}[/tex]

[tex]= 0.25[/tex]

Thus the above response is correct.

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