A) Step 1: Find change in the size of central diffraction band.
Formula used: y=Dλd
In a single slit diffraction experiment, the width of the central maxima is given by,
y=2Dλd
Where, λ= wavelength of light
D = distance between the slit and the screen
d = slit width.
If the width of the slit is doubled, then the size of the central diffraction band reduces to half.
Step 2: Find change in the intensity of central diffraction band.
Formula used: I∝d2
Intensity is directly proportional to the square of the slit width.
I∝d2
I=kd2
Here, k = constant
d = slit width
When the width of the slit is doubled then intensity becomes four times of the original intensity.
Final answer: Size reduces to half and intensity becomes four times.
B) In the double-slit experiment, the pattern on the screen is actually a superposition of single-slit diffraction from each slit. The interference pattern in a double slit experiment is modulated by diffraction from each slit.
Final answer: The intensity of interference fringes in a double slit arrangement is modulated by the diffraction pattern of each slit.
C) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
Final answer: Because of constructive interference.
D) As we know, bending of waves by obstacles by a large angle is possible only when the size of obstacle is comparable to the wavelength of the wave.
The wavelength of the light is too small in comparison to the size of the obstacle. So, the diffraction angle will be very small. Therefore, students are unable to see each other.
But the size of wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves take place at a large angle.
Here the size of obstacle is of the order of a few meters. The wavelength of light is about 5×10−7 m, while sound waves of, say, 1 kHz frequency have wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.
Therefore, students can hear each other.
Final answer: Due to the difference in their wavelengths.
E) In ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used. The wave effects of light like diffraction, interference etc. are not seen. Therefore, concept of ray optics is used in understanding location and several other properties of image in optical instruments.
Final answer: Size of aperture involved in ordinary optical instruments are much larger than the wavelength of the light.