Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Observations that indicate that visible light has a wavelength much shorter than a centimeter include which of the following?
I. The colored pattern seen in a soap bubble
II. The colored pattern seen when light passes through a diffraction grating
III. The bending of light when it passes from one medium to another medium
Correct Answer: \( \boxed{\mathrm{C}} \)
I. Soap bubble: The colors seen in a soap bubble are produced by thin-film interference. Interference occurs only when the film thickness is comparable to the wavelength of light, indicating that visible light has a very small wavelength. Therefore, statement I is correct.
II. Diffraction grating: A diffraction grating has extremely closely spaced slits. Significant diffraction and interference occur only when the slit spacing is comparable to the wavelength of light, demonstrating that visible light has a wavelength much smaller than a centimeter. Therefore, statement II is correct.
III. Refraction: Light bends when passing between media because its speed changes according to Snell’s law,
\( n_1\sin\theta_1=n_2\sin\theta_2. \)
Refraction occurs for waves of any wavelength and does not provide information about the magnitude of the wavelength. Therefore, statement III is not correct.
Hence, the observations that indicate visible light has a wavelength much shorter than a centimeter are I and II only.
As the number of lines per centimeter on a diffraction grating is increased (or the slit spacing is decreased in a two-slit diffraction pattern),
Correct Answer: \( \boxed{\mathrm{A}} \)
The condition for constructive interference from a diffraction grating is
\( d\sin\theta = m\lambda \)
where \(d\) is the spacing between adjacent slits, \(m\) is the order number, and \(\lambda\) is the wavelength of light.
Increasing the number of lines per centimeter decreases the slit spacing \(d\). For a given wavelength and diffraction order, a smaller \(d\) requires a larger value of \(\sin\theta\), so the diffraction angle \(\theta\) increases.
As the diffraction angles become larger, the bright spectral lines move farther apart on the screen, increasing the separation between adjacent spectral lines. This improves the ability to distinguish different wavelengths.
Therefore, the correct answer is (A).
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance \(D\) must be changed to
Correct Answer: \( \boxed{\mathrm{D}} \)
The fringe spacing in Young’s double-slit experiment is given by
\( x=\frac{\lambda D}{d} \)
where \(x\) is the fringe spacing, \(\lambda\) is the wavelength of the light, \(D\) is the distance from the slits to the screen, and \(d\) is the slit separation.
To keep the fringe spacing \(x\) constant, the ratio \(\frac{D}{d}\) must remain unchanged.
If the slit separation is doubled,
\( d’ = 2d \),
then the screen distance must also be doubled:
\( D’ = 2D \)
This keeps the ratio
\( \frac{D’}{d’}=\frac{2D}{2d}=\frac{D}{d} \),
so the fringe spacing remains unchanged.
Therefore, the correct answer is (D).
