1. If the wavelength of an electromagnetic wave is about the diameter of a cricket ball, what type of radiation is it?
Verification Matrix:
The diameter of a standard cricket ball is roughly $7.3\text{ cm}$ ($0.073\text{ m}$). In the electromagnetic spectrum, wavelengths ranging from a few centimeters up to thousands of kilometers fall squarely into the **Radio waves** category (specifically matching the short radio/microwave boundary region). X-rays ($10^{-10}\text{ m}$), UV ($10^{-8}\text{ m}$), and visible light ($10^{-7}\text{ m}$) are orders of magnitude smaller.
2. Electromagnetic waves from an unknown source in space are found to be diffracted when passing through gaps of the order of $10^{-5}\text{ m}$. Which type of wave are they most likely to be?
Verification Matrix:
Significant wave diffraction occurs only when the physical dimensions of the aperture or slit gap match the approximate order of magnitude of the wavelength ($\lambda \approx d$). A wavelength of $10^{-5}\text{ m}$ ($10\mu\text{m}$) falls explicitly within the **infrared** spectrum region, which spans roughly $700\text{ nm}$ up to $1\text{ mm}$.
3. Huygens's conception of secondary waves:
Verification Matrix:
Huygens' Principle states that every point on a primary wavefront acts as a source of secondary spherical wavelets. The forward envelope tangential to these secondary wavelets defines the new position of the progressive wavefront at any subsequent time interval. Thus, it serves primarily as a **geometrical method to locate subsequent wavefront shapes**.
4. Interference fringes are produced using monochromatic light of the same intensity from a double-slit screen. If the intensity of light emerging from one of the slits is reduced, the effect on the interference pattern will be:
Verification Matrix:
When both slits have equal wave amplitudes ($A_1 = A_2$), destructive interference is absolute: $I_{\text{min}} \propto (A_1 - A_2)^2 = 0$ (completely dark). If one slit's intensity is reduced, the amplitudes no longer match ($A_1 \neq A_2$). Consequently, destructive cancelation is incomplete, causing **dark fringes to become brighter** ($I_{\text{min}} > 0$). Concurrently, the total available system energy drops, meaning constructive maximum peaks drop, causing **bright fringes to become darker** ($I_{\text{max}} \propto (A_1 + A_2)^2$). This altogether degrades the pattern contrast.
5. In Young's experiment, when the distance between slits and screen is doubled while the separation of slits is halved, the fringe width will be:
Verification Matrix:
The analytical expression for fringe space width is $Y = \frac{\lambda L}{d}$. Substituting the new parameters ($L' = 2L$ and $d' = \frac{d}{2}$): $$Y' = \frac{\lambda (2L)}{(d / 2)} = 4 \left(\frac{\lambda L}{d}\right) = 4Y.$$
The spatial fringe spacing scales up by exactly **4 times**.
6. A ray of light passes from air into water, striking the surface with an angle of incidence of 45°. Which of these quantities change as the light enters the water?
i) Wavelength ii) Frequency iii) Speed of propagation iv) Direction of propagation
i) Wavelength ii) Frequency iii) Speed of propagation iv) Direction of propagation
Verification Matrix:
When entering a denser optical medium, light waves slow down ($v = \frac{c}{n}$) and bend towards the surface normal via refraction ($\theta_r < \theta_i$). Since the **frequency ($f$) depends strictly on the atomic source oscillator and remains constant**, the wavelength must shorten proportionally ($\lambda = \frac{v}{f}$). Therefore, parameters **i, iii, and iv change**, while frequency remains invariant.
7. A hill separates a television (TV) transmitter from a house. The transmitter cannot be seen from the house, but the TV still has good reception. What wave phenomenon makes this possible?
Verification Matrix:
The bending of long-wavelength radio and television signal waves around large physical barriers or geographical obstacles (like a hilltop) into shadow zones is a classic manifestation of **diffraction**.
8. Monochromatic light is incident on a diffraction grating and a pattern is observed. Which effect is observed by replacing the grating with one that has more lines per millimeter?
Verification Matrix:
The grating equation is $d \sin\theta = m\lambda$. Increasing lines per millimeter decreases the slit spacing ($d = \frac{1}{N}$). Because $\sin\theta = \frac{m\lambda}{d}$, decreasing $d$ forces $\sin\theta$ to grow, meaning the diffraction angles increase—**expanding the angular separation between successive orders**. Because the maximum possible physical angle is capped at $\theta = 90^\circ$, pushing the maxima further outward means fewer total orders fit within the physical space, so the **total number of visible maxima decreases**.
9. Optically active substances are those substances which:
Verification Matrix:
Optical activity is an asymmetric molecular property (like in sugar or quartz crystals) that causes a substance to **rotate the plane of polarization** of linearly polarized light passing through it, either clockwise (dextrorotatory) or counter-clockwise (levorotatory).
10. Plane-polarized light is passed through a Polaroid. On viewing through the Polaroid, we find that when the Polaroid is given one complete rotation about the direction of light:
Verification Matrix:
According to Malus's Law ($I = I_0 \cos^2\theta$), the transmitted light intensity depends on the relative angle between the light's polarization axis and the Polaroid's transmission axis. Over a full $360^\circ$ rotation, the term $\cos^2\theta$ reaches a value of $1$ twice (at $0^\circ$ and $180^\circ$) and drops to $0$ twice (at $90^\circ$ and $270^\circ$). This creates a pattern of **two maxima and two minima (zeroes)** per complete rotation.
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