When a monochromatic light wave passes from the air to glass across a plane surface of separation then:
A. its direction must change
B. its frequency must change
C. its wavelength increases
D. its wavelength decreases
Answer
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Hint: The passage of light from one medium to other medium across a surface of separation between them is called refraction. We will use the laws of refraction to predict the behavior of a light wave when it passes from the air to glass across a plane surface of separation.
Formula used:
Snell’s law-
${{n}_{1}}\sin {{\theta }_{1}}={{n}_{2}}\sin {{\theta }_{2}}$
Complete step by step answer:
We know from the laws of refraction that the light wave remains in the same plane and from Snell's law formula we can see that ${{n}_{1}}\sin {{\theta }_{1}}={{n}_{2}}\sin {{\theta }_{2}}$, where, ${{n}_{1}}$ is the refractive index of the first medium which is air in this case through which the light wave passes and ${{\theta }_{1}}$ is the angle the light wave makes with the normal at the point of contact with the surface boundary and ${{n}_{2}}$ is the refractive index of the second medium which is glass in this case, through which the light wave passes and \[{{\theta }_{2}}\] is the angle the light wave makes with the normal at the point of emergence from the surface boundary.
Can there be some value of the ${{\theta }_{1}}$ for which ${{\theta }_{2}}$ is the same as ${{\theta }_{1}}$? We will observe that when the light wave passes through the surface boundary along the normal, its direction of motion does not change. Hence the first option is incorrect. Next, we know that the frequency of the waves does not change on changing the medium. However, as the speed of light changes in glass and the frequency remains constant, the wavelength of the light wave must change. As the speed is a product of frequency and wavelength and the speed decreases in glass, the wavelength must also decrease. Hence, the correct option is D, i.e. its wavelength decreases.
So, the correct answer is “Option D”.
Note: Students must take care that the frequency of a wave changes only when the frequency of the driving force or frequency of the source changes. There are no changes in frequency due to reflection or refraction. The wavelength however may change.
Formula used:
Snell’s law-
${{n}_{1}}\sin {{\theta }_{1}}={{n}_{2}}\sin {{\theta }_{2}}$
Complete step by step answer:
We know from the laws of refraction that the light wave remains in the same plane and from Snell's law formula we can see that ${{n}_{1}}\sin {{\theta }_{1}}={{n}_{2}}\sin {{\theta }_{2}}$, where, ${{n}_{1}}$ is the refractive index of the first medium which is air in this case through which the light wave passes and ${{\theta }_{1}}$ is the angle the light wave makes with the normal at the point of contact with the surface boundary and ${{n}_{2}}$ is the refractive index of the second medium which is glass in this case, through which the light wave passes and \[{{\theta }_{2}}\] is the angle the light wave makes with the normal at the point of emergence from the surface boundary.
Can there be some value of the ${{\theta }_{1}}$ for which ${{\theta }_{2}}$ is the same as ${{\theta }_{1}}$? We will observe that when the light wave passes through the surface boundary along the normal, its direction of motion does not change. Hence the first option is incorrect. Next, we know that the frequency of the waves does not change on changing the medium. However, as the speed of light changes in glass and the frequency remains constant, the wavelength of the light wave must change. As the speed is a product of frequency and wavelength and the speed decreases in glass, the wavelength must also decrease. Hence, the correct option is D, i.e. its wavelength decreases.
So, the correct answer is “Option D”.
Note: Students must take care that the frequency of a wave changes only when the frequency of the driving force or frequency of the source changes. There are no changes in frequency due to reflection or refraction. The wavelength however may change.
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