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In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:
- a)3.9 mm
- b)6.9 mm
- c)5.9 mm
- d)4.9 mm
Correct answer is option 'C'. Can you explain this answer?
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In a Youngs double slit experiment, the separation between the slits i...
To find the separation between the successive bright fringes on the screen in a Young's double slit experiment, we can use the formula:
δy = λL/d
Where:
δy is the separation between the bright fringes on the screen,
λ is the wavelength of light used,
L is the distance between the double slits and the screen,
and d is the separation between the double slits.
Given:
λ = 589 nm (or 589 × 10^-9 m),
L = 1.5 m,
and d = 0.15 mm (or 0.15 × 10^-3 m).
Let's substitute these values into the formula to find the separation between the bright fringes:
δy = (589 × 10^-9 m) × (1.5 m) / (0.15 × 10^-3 m)
= (589 × 1.5) / 0.15 × 10^-6
= 883.5 × 10^-6 / 0.15 × 10^-6
= 883.5 / 0.15
≈ 5890 mm
Therefore, the separation between the successive bright fringes on the screen is approximately 5890 mm.
Now let's convert this to the given options:
a) 3.9 mm
b) 6.9 mm
c) 5.9 mm
d) 4.9 mm
We can see that the correct answer is option c) 5.9 mm.
Therefore, the separation between the successive bright fringes on the screen is 5.9 mm.
δy = λL/d
Where:
δy is the separation between the bright fringes on the screen,
λ is the wavelength of light used,
L is the distance between the double slits and the screen,
and d is the separation between the double slits.
Given:
λ = 589 nm (or 589 × 10^-9 m),
L = 1.5 m,
and d = 0.15 mm (or 0.15 × 10^-3 m).
Let's substitute these values into the formula to find the separation between the bright fringes:
δy = (589 × 10^-9 m) × (1.5 m) / (0.15 × 10^-3 m)
= (589 × 1.5) / 0.15 × 10^-6
= 883.5 × 10^-6 / 0.15 × 10^-6
= 883.5 / 0.15
≈ 5890 mm
Therefore, the separation between the successive bright fringes on the screen is approximately 5890 mm.
Now let's convert this to the given options:
a) 3.9 mm
b) 6.9 mm
c) 5.9 mm
d) 4.9 mm
We can see that the correct answer is option c) 5.9 mm.
Therefore, the separation between the successive bright fringes on the screen is 5.9 mm.
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Community Answer
In a Youngs double slit experiment, the separation between the slits i...
Fringe-width,
= 589 × 10-2 mm = 5.89 mm
≈ 5.9 mm
≈ 5.9 mm
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In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer?
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