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In Young's double slit experiment, the width of the fringes obtained with light of wavelength 6000 Angstrom is 2.0mm. The fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33, will be
  • a)
    1.5m
  • b)
    1.5cm
  • c)
    1.5mm
  • d)
    1.5km
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In Young's double slit experiment, the width of the fringes obtain...
Fringe width in Young's double slit experiment β = Dλ/d
When apparatus is immersed in liquid, only wavelength of light (λ) changes
Wavelength of light is liquid,
λ' = λ/n , where n = refractive index of medium
Initial fringe width, β = Dλ/d .....(i)
Fringe width in liquid, β' = Dλ'/d .....(ii)
Dividing Equation (ii) by Equation (i), we get
β'/β = λ'/λ
or β'/β = λ n/λ = 1/n
or β' = β/n
The new fringe width, β' = 2.0/1.33 mm = 1.5 mm
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Most Upvoted Answer
In Young's double slit experiment, the width of the fringes obtain...
Young's double-slit experiment involves passing light through two narrow slits and observing the interference pattern formed on a screen. The interference pattern consists of bright and dark fringes. The width of these fringes can be determined using the formula:

w = λL / d

where w is the fringe width, λ is the wavelength of light, L is the distance between the double-slit and the screen, and d is the distance between the two slits.

In this case, the given wavelength of light is 6000 Angstrom (600 nm), and the fringe width is 2.0 mm.

1. Calculate the fringe width without the liquid
Using the formula, we can rearrange it to solve for L:

L = w * d / λ

Substituting the given values, we get:

L = (2.0 mm) * d / 600 nm

2. Calculate the refractive index of the liquid
The refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:

n = c / v

where c is the speed of light in vacuum and v is the speed of light in the medium.

In this case, the refractive index of the liquid is given as 1.33.

3. Calculate the fringe width in the liquid
When the entire apparatus is immersed in the liquid, the wavelength of light in the liquid changes. The new wavelength (λ') can be calculated using the equation:

λ' = λ / n

Substituting the values, we get:

λ' = (600 nm) / 1.33

Now, we can calculate the new fringe width (w') using the formula:

w' = λ'L / d

Substituting the values, we get:

w' = [(600 nm) / 1.33] * L / d

4. Compare the fringe widths
To compare the fringe widths, we can take the ratio of the two widths:

w' / w = [(600 nm) / 1.33] * L / d / (2.0 mm) * d / 600 nm

Simplifying the expression, we get:

w' / w = 1 / 1.33

Therefore, the new fringe width (w') is 1.33 times smaller than the original fringe width (w).

The correct answer is option C. The fringe width will be 1.5 mm.
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In Young's double slit experiment, the width of the fringes obtained with light of wavelength 6000 Angstrom is 2.0mm. The fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33, will bea)1.5mb)1.5cmc)1.5mmd)1.5kmCorrect answer is option 'C'. Can you explain this answer?
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