
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. Find the value of 'x' to the nearest integer.
Answer
163.8k+ views
Hint: To solve this problem we have to use the concept of Young's double slit experiment and formula of bandwidth and get the value of x. Young's double slit experiment is a distinctive pattern of bright and dark fringes as seen when monochromatic light illuminates a distant screen after passing through two small openings. The superposition of overlapping light waves coming from the two slits results in this interference pattern.
Formula used:
The expression of fringe width ($\beta$) is,
$\beta = \dfrac{{\lambda D}}{d}$
Here, $\lambda$ is the wavelength of monochromatic light, $d$ is the slit separation distance and $D$ is the distance between silt and screen.
Complete step by step solution:
The distance between two subsequent brilliant fringes or two subsequent dark fringes is known as the fringe width. The interference pattern's fringes all have the same width of fringe. The wavelength of the light being used is directly related to the fringe width. According to question, it is given that;
\[\beta = 6{\text{ }}mm \\
\Rightarrow d = 1{\text{ }}mm \\
\Rightarrow D = 10{\text{ }}m \\ \]
Now putting the given values in the fringe width expression and we get the wavelength as,
\[\beta = \dfrac{{\lambda D}}{d} \\
\Rightarrow 6 \times {10^{ - 3}} = \dfrac{{\lambda \times 10}}{{1 \times {{10}^{ - 3}}}} \\
\Rightarrow \lambda = \dfrac{{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}}}{{10}} \\
\Rightarrow \lambda = 600{\text{ }} \times {\text{ }}{10^{ - 9}}\;m \\
\therefore \lambda {\text{ }} = {\text{ }}600{\text{ }}nm \\ \]
Hence, the value of ‘x’ is \[600\].
Note: Two coherent sources of light are employed in Young's double slit experiment, which is often conducted at a distance that is only a few orders of magnitude higher than the wavelength of the light used. Young's double slit experiment contributed to our knowledge of the diagrammed wave theory of light. The space between two consecutive bright spots (maximas, when constructive interference occurs), or two consecutive dark patches, is known as the fringe width (minimas, where destructive interference takes place).
Formula used:
The expression of fringe width ($\beta$) is,
$\beta = \dfrac{{\lambda D}}{d}$
Here, $\lambda$ is the wavelength of monochromatic light, $d$ is the slit separation distance and $D$ is the distance between silt and screen.
Complete step by step solution:
The distance between two subsequent brilliant fringes or two subsequent dark fringes is known as the fringe width. The interference pattern's fringes all have the same width of fringe. The wavelength of the light being used is directly related to the fringe width. According to question, it is given that;
\[\beta = 6{\text{ }}mm \\
\Rightarrow d = 1{\text{ }}mm \\
\Rightarrow D = 10{\text{ }}m \\ \]
Now putting the given values in the fringe width expression and we get the wavelength as,
\[\beta = \dfrac{{\lambda D}}{d} \\
\Rightarrow 6 \times {10^{ - 3}} = \dfrac{{\lambda \times 10}}{{1 \times {{10}^{ - 3}}}} \\
\Rightarrow \lambda = \dfrac{{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}}}{{10}} \\
\Rightarrow \lambda = 600{\text{ }} \times {\text{ }}{10^{ - 9}}\;m \\
\therefore \lambda {\text{ }} = {\text{ }}600{\text{ }}nm \\ \]
Hence, the value of ‘x’ is \[600\].
Note: Two coherent sources of light are employed in Young's double slit experiment, which is often conducted at a distance that is only a few orders of magnitude higher than the wavelength of the light used. Young's double slit experiment contributed to our knowledge of the diagrammed wave theory of light. The space between two consecutive bright spots (maximas, when constructive interference occurs), or two consecutive dark patches, is known as the fringe width (minimas, where destructive interference takes place).
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