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In young's experiment, two coherent sources are 0.90 mm apart and fringes are observed at a distance of 1 m, if 2nd dark fringe is at 1 mm distance from central fringe, then wavelength of the monochromatic light will be
- a)60 x 10-4 cm
- b)10 x 10-4 cm
- c)10 x 10-5 cm
- d)6 x 10-5 cm
Correct answer is option 'D'. Can you explain this answer?
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In youngs experiment, two coherent sources are 0.90 mm apart and fring...
d = 0.90 x 10-3 M, D = 1 m
λ = 6 x 10-5 cm
λ = 6 x 10-5 cm
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In youngs experiment, two coherent sources are 0.90 mm apart and fring...
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In youngs experiment, two coherent sources are 0.90 mm apart and fring...
To solve this problem, we can use the formula for the fringe width in Young's experiment:
\(w = \frac{{\lambda D}}{{d}}\)
where:
\(w\) is the fringe width,
\(\lambda\) is the wavelength of the light,
\(D\) is the distance between the screen and the sources, and
\(d\) is the distance between the two sources.
We are given that the distance between the two coherent sources is 0.90 mm and the distance between the screen and the sources is 1 m.
Let's first calculate the fringe width using the given data:
\(w = \frac{{\lambda D}}{{d}}\)
\(w = \frac{{\lambda \times 1}}{{0.90 \times 10^{-3}}}\)
Now, we are also given that the 2nd dark fringe is at a distance of 1 mm from the central fringe. Since the dark fringes occur at odd multiples of the fringe width, the 2nd dark fringe is at a distance of \(2w\) from the central fringe.
Given that the 2nd dark fringe is at a distance of 1 mm, we can calculate the fringe width using the formula:
\(2w = 1 \times 10^{-3}\)
Now we have two equations with two unknowns (\(\lambda\) and \(w\)). We can solve these equations simultaneously to find the value of \(\lambda\).
From the equation \(2w = 1 \times 10^{-3}\), we can solve for \(w\) and substitute it into the other equation:
\(w = \frac{{\lambda \times 1}}{{0.90 \times 10^{-3}}}\)
\(w = \frac{{1 \times 10^{-3}}}{2}\)
Substituting this value of \(w\) into the other equation, we get:
\(\frac{{1 \times 10^{-3}}}{2} = \frac{{\lambda \times 1}}{{0.90 \times 10^{-3}}}\)
Simplifying the equation, we find:
\(\lambda = \frac{{1 \times 10^{-3}} \times 0.90 \times 10^{-3}}{2 \times 1}\)
\(\lambda = 0.45 \times 10^{-6}\)
Converting this to centimeters, we get:
\(\lambda = 4.5 \times 10^{-7}\) cm
Hence, the correct answer is option D: 6 x 10^-5 cm.
\(w = \frac{{\lambda D}}{{d}}\)
where:
\(w\) is the fringe width,
\(\lambda\) is the wavelength of the light,
\(D\) is the distance between the screen and the sources, and
\(d\) is the distance between the two sources.
We are given that the distance between the two coherent sources is 0.90 mm and the distance between the screen and the sources is 1 m.
Let's first calculate the fringe width using the given data:
\(w = \frac{{\lambda D}}{{d}}\)
\(w = \frac{{\lambda \times 1}}{{0.90 \times 10^{-3}}}\)
Now, we are also given that the 2nd dark fringe is at a distance of 1 mm from the central fringe. Since the dark fringes occur at odd multiples of the fringe width, the 2nd dark fringe is at a distance of \(2w\) from the central fringe.
Given that the 2nd dark fringe is at a distance of 1 mm, we can calculate the fringe width using the formula:
\(2w = 1 \times 10^{-3}\)
Now we have two equations with two unknowns (\(\lambda\) and \(w\)). We can solve these equations simultaneously to find the value of \(\lambda\).
From the equation \(2w = 1 \times 10^{-3}\), we can solve for \(w\) and substitute it into the other equation:
\(w = \frac{{\lambda \times 1}}{{0.90 \times 10^{-3}}}\)
\(w = \frac{{1 \times 10^{-3}}}{2}\)
Substituting this value of \(w\) into the other equation, we get:
\(\frac{{1 \times 10^{-3}}}{2} = \frac{{\lambda \times 1}}{{0.90 \times 10^{-3}}}\)
Simplifying the equation, we find:
\(\lambda = \frac{{1 \times 10^{-3}} \times 0.90 \times 10^{-3}}{2 \times 1}\)
\(\lambda = 0.45 \times 10^{-6}\)
Converting this to centimeters, we get:
\(\lambda = 4.5 \times 10^{-7}\) cm
Hence, the correct answer is option D: 6 x 10^-5 cm.
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In youngs experiment, two coherent sources are 0.90 mm apart and fringes are observed at a distance of 1 m, if 2nd dark fringe is at 1 mm distance from central fringe, then wavelength of the monochromatic light will bea)60 x 10-4 cmb)10 x 10-4cmc)10 x 10-5cmd)6 x 10-5cmCorrect answer is option 'D'. Can you explain this answer?
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