
If the wavelength of light is \[4000\mathop A\limits^ \circ \], then the number of waves in 1 mm length will be
A. 2500
B. 25
C. 250
D. 25000
Answer
164.1k+ views
Hint:The horizontal distance of a one complete wave cycle is called the wavelength of the wave. Then the total horizontal length along the propagation of the wave is an integral multiple of the wavelength.
Formula used:
\[L = n\lambda \]
where L is the total horizontal length along the wave propagation, n is the number of waves and \[\lambda \] is the wavelength of the wave.
Complete step by step solution:
We have given the wavelength of the wave and need to find the number of waves present in a particular length. The horizontal distance of a complete wave cycle is called the wavelength of the wave. A complete wave contains one crest and one trough. Then the total horizontal length along the propagation of the wave is an integral multiple of the wavelength.
It is given that the wavelength of the wave is \[4000\mathop A\limits^ \circ \]. We need to convert the given wavelength in S.I. unit. As we know that 1 angstrom is equal to \[{10^{ - 10}}m\], hence the wavelength will be;
\[\lambda = 4000 \times {10^{ - 10}}m\]
\[\Rightarrow \lambda = 4.0 \times {10^{ - 7}}m\]
The horizontal length is given as 1 mm.
\[L = {10^{ - 3}}m\]
We need to find the number of waves in the length of 1 mm, i.e. \[{10^{ - 3}}m\]
Using the formula of the wave number, we get
\[n = \dfrac{L}{\lambda }\]
\[\Rightarrow n = \dfrac{{{{10}^{ - 3}}}}{{4 \times {{10}^{ - 7}}}}\]
\[\therefore n = 2500\]
Hence, there are a total of 2500 complete waves present in the length of 1 mm.
Therefore, the correct option is A.
Note: We should be careful about the term wave number and the number of waves. The wave number is the characteristic feature of the wave and is constant for a wave of a particular wavelength, as it is given by the expression\[K = \dfrac{{2\pi }}{\lambda }\], whereas the number of waves is the total number of waves which can be accumulated in a particular horizontal length.
Formula used:
\[L = n\lambda \]
where L is the total horizontal length along the wave propagation, n is the number of waves and \[\lambda \] is the wavelength of the wave.
Complete step by step solution:
We have given the wavelength of the wave and need to find the number of waves present in a particular length. The horizontal distance of a complete wave cycle is called the wavelength of the wave. A complete wave contains one crest and one trough. Then the total horizontal length along the propagation of the wave is an integral multiple of the wavelength.
It is given that the wavelength of the wave is \[4000\mathop A\limits^ \circ \]. We need to convert the given wavelength in S.I. unit. As we know that 1 angstrom is equal to \[{10^{ - 10}}m\], hence the wavelength will be;
\[\lambda = 4000 \times {10^{ - 10}}m\]
\[\Rightarrow \lambda = 4.0 \times {10^{ - 7}}m\]
The horizontal length is given as 1 mm.
\[L = {10^{ - 3}}m\]
We need to find the number of waves in the length of 1 mm, i.e. \[{10^{ - 3}}m\]
Using the formula of the wave number, we get
\[n = \dfrac{L}{\lambda }\]
\[\Rightarrow n = \dfrac{{{{10}^{ - 3}}}}{{4 \times {{10}^{ - 7}}}}\]
\[\therefore n = 2500\]
Hence, there are a total of 2500 complete waves present in the length of 1 mm.
Therefore, the correct option is A.
Note: We should be careful about the term wave number and the number of waves. The wave number is the characteristic feature of the wave and is constant for a wave of a particular wavelength, as it is given by the expression\[K = \dfrac{{2\pi }}{\lambda }\], whereas the number of waves is the total number of waves which can be accumulated in a particular horizontal length.
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