301. In a single-slit diffraction pattern, the first minimum is observed at a distance \(y\) from the central point on a screen at distance \(D\). For small angles, the slit width \(a\) is approximately
ⓐ. \(\frac{\lambda y}{D}\)
ⓑ. \(\frac{D}{\lambda y}\)
ⓒ. \(\frac{yD}{\lambda}\)
ⓓ. \(\frac{\lambda D}{y}\)
Correct Answer: \(\frac{\lambda D}{y}\)
Explanation: The first minimum in single-slit diffraction satisfies \(a\sin\theta=\lambda\).
For small angles, \(\sin\theta\approx\tan\theta\approx\frac{y}{D}\).
Substituting this into the first-minimum condition gives
\[
a\frac{y}{D}=\lambda
\]
Solving for slit width gives
\[
a=\frac{\lambda D}{y}
\]
The result shows that a larger distance of the first minimum from the centre corresponds to a narrower slit for the same \(D\) and \(\lambda\). The expression \(\frac{\lambda y}{D}\) reverses the geometrical relation between screen displacement and angle.
302. A single slit produces its first minimum at \(2.0\,\text{mm}\) from the centre on a screen \(1.0\,\text{m}\) away. If the wavelength is \(500\,\text{nm}\), the slit width is
ⓐ. \(0.25\,\text{mm}\)
ⓑ. \(0.50\,\text{mm}\)
ⓒ. \(0.10\,\text{mm}\)
ⓓ. \(1.00\,\text{mm}\)
Correct Answer: \(0.25\,\text{mm}\)
Explanation: \( \textbf{Given:} \) \(y=2.0\,\text{mm}\), \(D=1.0\,\text{m}\), and \(\lambda=500\,\text{nm}\).
\( \textbf{Required:} \) Slit width \(a\).
\( \textbf{First-minimum condition:} \)
\[
a\sin\theta=\lambda
\]
\( \textbf{Small-angle step:} \)
\[
\sin\theta\approx\frac{y}{D}
\]
\( \textbf{So:} \)
\[
a\frac{y}{D}=\lambda
\]
\( \textbf{Rearrange:} \)
\[
a=\frac{\lambda D}{y}
\]
\( \textbf{Unit conversion:} \)
\[
\lambda=500\,\text{nm}=5.0\times10^{-7}\,\text{m}
\]
\[
y=2.0\,\text{mm}=2.0\times10^{-3}\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
a=\frac{(5.0\times10^{-7})(1.0)}{2.0\times10^{-3}}\,\text{m}
\]
\( \textbf{Calculation:} \)
\[
a=2.5\times10^{-4}\,\text{m}=0.25\,\text{mm}
\]
\( \textbf{Final answer:} \) The slit width is \(0.25\,\text{mm}\).
303. A graph is plotted with the angular width of the central maximum on the vertical axis and \(\frac{1}{a}\) on the horizontal axis for a single-slit diffraction experiment. If \(\lambda\) is fixed, the slope of the graph is
ⓐ. \(\lambda\)
ⓑ. \(2\lambda\)
ⓒ. \(\frac{a}{2\lambda}\)
ⓓ. \(\frac{1}{2\lambda}\)
Correct Answer: \(2\lambda\)
Explanation: The angular width of the central maximum in single-slit diffraction is approximately \(\frac{2\lambda}{a}\).
Writing it in terms of the horizontal variable \(\frac{1}{a}\), we get
\[
\text{angular width}=2\lambda\left(\frac{1}{a}\right)
\]
This is a straight-line relation of the form \(y=mx\). Therefore, the slope is \(2\lambda\). The graph passes through the origin in the ideal small-angle model. The factor \(2\) appears because the central maximum extends from the first minimum on one side to the first minimum on the other side.
304. A table compares single-slit diffraction quantities.
| Row | Quantity | Expression or meaning |
|---|
| P | First minimum | \(a\sin\theta=\lambda\) |
| Q | Angular half-width of central maximum | \(\frac{\lambda}{a}\) for small angles |
| R | Angular full width of central maximum | \(\frac{2\lambda}{a}\) for small angles |
| S | Linear full width on screen | \(\frac{\lambda D}{2a}\) |
The row that should be corrected is
ⓐ. Row R
ⓑ. Row P
ⓒ. Row Q
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P is correct because the first minimum of a single slit satisfies \(a\sin\theta=\lambda\). Row Q is correct because the angular half-width is the angle from the centre to the first minimum, approximately \(\frac{\lambda}{a}\). Row R is correct because the central maximum extends to the first minimum on both sides, giving full angular width \(\frac{2\lambda}{a}\). Row S is wrong because the linear full width on a screen at distance \(D\) is \(\frac{2\lambda D}{a}\), not \(\frac{\lambda D}{2a}\). The factor of \(2\) belongs in the numerator for full width.
305. In a single-slit diffraction pattern, the second minimum occurs at a direction where
ⓐ. \(a\sin\theta=3\lambda\)
ⓑ. \(a\sin\theta=\frac{\lambda}{2}\)
ⓒ. \(a\sin\theta=\lambda\)
ⓓ. \(a\sin\theta=2\lambda\)
Correct Answer: \(a\sin\theta=2\lambda\)
Explanation: For a single slit, minima occur at \(a\sin\theta=n\lambda\), where \(n=1,2,3,\ldots\). The first minimum corresponds to \(n=1\). The second minimum corresponds to \(n=2\). Therefore, the condition for the second minimum is \(a\sin\theta=2\lambda\). The condition \(a\sin\theta=\frac{\lambda}{2}\) is not a single-slit minimum condition; it resembles a two-source half-path condition.
306. In single-slit diffraction, if the first minimum occurs at angle \(\theta_1\) and the second minimum at angle \(\theta_2\), then for small angles
ⓐ. \(\theta_2\approx4\theta_1\)
ⓑ. \(\theta_2\approx2\theta_1\)
ⓒ. \(\theta_2\approx\frac{\theta_1}{2}\)
ⓓ. \(\theta_2\approx\theta_1\)
Correct Answer: \(\theta_2\approx2\theta_1\)
Explanation: \( \textbf{First minimum:} \)
\[
a\sin\theta_1=\lambda
\]
\( \textbf{Small-angle form:} \)
\[
a\theta_1\approx\lambda
\]
\( \textbf{So:} \)
\[
\theta_1\approx\frac{\lambda}{a}
\]
\( \textbf{Second minimum:} \)
\[
a\sin\theta_2=2\lambda
\]
\( \textbf{Small-angle form:} \)
\[
a\theta_2\approx2\lambda
\]
\( \textbf{So:} \)
\[
\theta_2\approx\frac{2\lambda}{a}
\]
\( \textbf{Comparison:} \)
\[
\theta_2\approx2\theta_1
\]
\( \textbf{Final answer:} \) The second minimum is approximately at twice the angle of the first minimum.
307. Use the intensity pattern described below.
A single-slit diffraction pattern has a central maximum at \(O\). The first minima are at points \(P\) and \(Q\) on opposite sides of \(O\). The second minima are farther out on both sides.
The width \(PQ\) represents
ⓐ. the wavelength \(\lambda\) directly
ⓑ. the distance between two consecutive secondary maxima
ⓒ. the linear width of the central maximum
ⓓ. the slit width \(a\)
Correct Answer: the linear width of the central maximum
Explanation: In a single-slit diffraction pattern, the central maximum extends between the first minima on the two sides. If \(P\) and \(Q\) are the first minima on opposite sides of the central point \(O\), then the distance \(PQ\) is the full linear width of the central maximum. The slit width \(a\) is the physical width of the aperture, not a distance measured on the screen. The wavelength affects the value of \(PQ\), but \(PQ\) is not itself equal to \(\lambda\). This interpretation connects the screen pattern with the minima condition.
308. A single slit is illuminated by light of wavelength \(\lambda\). If the entire setup is immersed in a liquid where the wavelength becomes \(\frac{\lambda}{n}\), while slit width \(a\) remains unchanged, the angular width of the central maximum becomes
ⓐ. unchanged from its air value
ⓑ. multiplied by \(n\)
ⓒ. divided by \(n\)
ⓓ. multiplied by \(n^2\)
Correct Answer: divided by \(n\)
Explanation: The angular width of the central maximum is approximately \(\frac{2\lambda}{a}\), where \(\lambda\) is the wavelength in the medium. In the liquid, the wavelength becomes \(\frac{\lambda}{n}\). Substituting this into the expression gives
\[
\text{new angular width}=\frac{2(\lambda/n)}{a}
\]
This is \(\frac{1}{n}\) times the original value \(\frac{2\lambda}{a}\). The slit width does not change, so the change comes only from the reduced wavelength in the medium. A smaller wavelength produces less angular spreading for the same slit width.
309. A claim says, “In single-slit diffraction, a wider central maximum means a wider physical slit.” The better evaluation is that
ⓐ. correct, because width is proportional to slit width
ⓑ. incorrect; width varies inversely with slit width
ⓒ. correct only when the wavelength is zero
ⓓ. unrelated, because diffraction ignores aperture size
Correct Answer: incorrect; width varies inversely with slit width
Explanation: The central maximum width is controlled by \(\frac{2\lambda D}{a}\) on a distant screen. For fixed wavelength \(\lambda\) and screen distance \(D\), the width is inversely proportional to slit width \(a\). A narrower slit produces a wider diffraction pattern. A wider slit gives less spreading and a narrower central maximum. The screen pattern width and the physical slit width therefore vary oppositely in this diffraction situation.
310. The resolving power of an optical instrument is closely connected with diffraction because
ⓐ. interference is absent in all optical instruments
ⓑ. apertures form diffraction patterns
ⓒ. wavelength has no role in instrument images
ⓓ. diffraction makes every image perfectly sharp
Correct Answer: apertures form diffraction patterns
Explanation: An optical instrument has a finite aperture, so light from a point object does not form a perfect geometrical point image. Because of diffraction, it forms a central bright disc with surrounding rings. When two point objects are close together, their diffraction patterns may overlap. The ability to distinguish them as separate objects is therefore limited by diffraction. This is why resolving power depends on aperture size and wavelength rather than on ray optics alone.
311. According to the Rayleigh criterion, two point images are just resolved when
ⓐ. the first maxima of both images vanish
ⓑ. the two central maxima exactly coincide
ⓒ. one maximum falls on the first minimum
ⓓ. the aperture diameter becomes zero
Correct Answer: one maximum falls on the first minimum
Explanation: The Rayleigh criterion gives a practical condition for just resolving two close point sources. Each point source forms a diffraction pattern due to the finite aperture of the instrument. The two sources are just resolved when the central maximum of one pattern coincides with the first minimum of the other pattern. At this condition, the combined pattern has enough dip between the two peaks to distinguish them. If the central maxima coincide, the two sources are not resolved as separate.
312. For a telescope with a circular objective of diameter \(D_o\), the minimum angular separation that can be resolved is approximately
ⓐ. \(\theta_{\min}=\frac{1.22\lambda}{D_o}\)
ⓑ. \(\theta_{\min}=\frac{D_o}{1.22\lambda}\)
ⓒ. \(\theta_{\min}=1.22\lambda D_o\)
ⓓ. \(\theta_{\min}=\frac{\lambda D_o}{1.22}\)
Correct Answer: \(\theta_{\min}=\frac{1.22\lambda}{D_o}\)
Explanation: A circular aperture produces an Airy diffraction pattern. The angular radius of the first dark ring is approximately \(\frac{1.22\lambda}{D_o}\). By the Rayleigh criterion, this angle is used as the minimum angular separation for just resolving two point objects. Therefore, \(\theta_{\min}=\frac{1.22\lambda}{D_o}\). A larger objective diameter reduces this angle and improves resolution. A shorter wavelength also improves resolving ability because the diffraction pattern becomes narrower.
313. A telescope objective diameter is doubled while the wavelength of light remains the same. The minimum resolvable angular separation becomes
ⓐ. unchanged
ⓑ. double
ⓒ. four times
ⓓ. half
Correct Answer: half
Explanation: The diffraction-limited angular resolution of a circular telescope aperture is approximately \(\theta_{\min}=\frac{1.22\lambda}{D_o}\). If the objective diameter \(D_o\) is doubled, the denominator becomes twice as large. Therefore, \(\theta_{\min}\) becomes half of its earlier value. A smaller minimum resolvable angle means the instrument can distinguish closer angular details. This is why larger apertures improve resolving power.
314. A telescope objective has diameter \(0.10\,\text{m}\). It observes light of wavelength \(500\,\text{nm}\). The approximate minimum angular separation for resolution is
ⓐ. \(6.1\times10^{-6}\,\text{rad}\)
ⓑ. \(2.44\times10^{-3}\,\text{rad}\)
ⓒ. \(6.1\times10^{-5}\,\text{rad}\)
ⓓ. \(1.22\times10^{-4}\,\text{rad}\)
Correct Answer: \(6.1\times10^{-6}\,\text{rad}\)
Explanation: \( \textbf{Given:} \) \(D_o=0.10\,\text{m}\) and \(\lambda=500\,\text{nm}\).
\( \textbf{Required:} \) Minimum angular separation \(\theta_{\min}\).
\( \textbf{Resolution relation for circular aperture:} \)
\[
\theta_{\min}=\frac{1.22\lambda}{D_o}
\]
\( \textbf{Unit conversion:} \)
\[
500\,\text{nm}=5.00\times10^{-7}\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
\theta_{\min}=\frac{1.22(5.00\times10^{-7})}{0.10}\,\text{rad}
\]
\( \textbf{Numerator:} \)
\[
1.22(5.00\times10^{-7})=6.10\times10^{-7}
\]
\( \textbf{Division:} \)
\[
\theta_{\min}=6.10\times10^{-6}\,\text{rad}
\]
\( \textbf{Final answer:} \) The minimum resolvable angular separation is \(6.1\times10^{-6}\,\text{rad}\).
315. A table describes telescope resolution.
| Row | Change | Effect on \(\theta_{\min}\) |
|---|
| P | \(\lambda\) decreased, \(D_o\) fixed | \(\theta_{\min}\) decreases |
| Q | \(D_o\) increased, \(\lambda\) fixed | \(\theta_{\min}\) decreases |
| R | \(\lambda\) increased, \(D_o\) fixed | \(\theta_{\min}\) increases |
| S | \(D_o\) increased, \(\lambda\) fixed | \(\theta_{\min}\) increases |
The row that should be revised is
ⓐ. Row Q
ⓑ. Row P
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: For a circular telescope aperture, \(\theta_{\min}=\frac{1.22\lambda}{D_o}\). Row P is correct because decreasing wavelength reduces the minimum resolvable angle. Row Q is correct because increasing aperture diameter also reduces \(\theta_{\min}\). Row R is correct because using a longer wavelength increases diffraction spreading and worsens angular resolution. Row S contradicts the inverse dependence on \(D_o\). A larger objective improves resolution by making \(\theta_{\min}\) smaller.
316. Resolving power of a telescope improves when the minimum resolvable angular separation
ⓐ. becomes equal to wavelength in metres
ⓑ. increases
ⓒ. decreases
ⓓ. becomes independent of aperture
Correct Answer: decreases
Explanation: Resolving power describes the ability to distinguish two close objects as separate. If the minimum angular separation needed for resolution is smaller, the instrument can separate objects that are closer together in angle. Therefore, smaller \(\theta_{\min}\) means better resolving ability. Since \(\theta_{\min}\approx\frac{1.22\lambda}{D_o}\), increasing aperture or decreasing wavelength improves the resolving power. The value is an angle, not a length equal to the wavelength.
317. A microscope objective has numerical aperture \(n\sin\alpha\). The resolving power is improved by
ⓐ. increasing \(n\sin\alpha\) or decreasing \(\lambda\)
ⓑ. using only longer wavelength light
ⓒ. decreasing \(n\sin\alpha\) and increasing \(\lambda\)
ⓓ. making the aperture zero
Correct Answer: increasing \(n\sin\alpha\) or decreasing \(\lambda\)
Explanation: In a microscope, the smallest resolvable distance is reduced when the numerical aperture is increased. The numerical aperture is \(n\sin\alpha\), where \(n\) is the refractive index of the medium and \(\alpha\) is the semi-angle of the cone of light entering the objective. A larger numerical aperture allows more diffracted light to be collected. A shorter wavelength also reduces diffraction spreading. Both changes make it easier to distinguish closely spaced details.
318. A microscope is used first with air and then with an oil-immersion medium of higher refractive index, keeping the objective geometry suitable. The resolution improves mainly because
ⓐ. diffraction completely disappears
ⓑ. the object becomes self-luminous
ⓒ. the numerical aperture increases
ⓓ. wavelength becomes irrelevant
Correct Answer: the numerical aperture increases
Explanation: In microscope resolution, the numerical aperture \(n\sin\alpha\) is important. Using an immersion oil with higher refractive index can increase \(n\), and it can also help collect a wider cone of light. This increases the numerical aperture and improves the ability to resolve fine details. Diffraction does not disappear, but its limiting effect is reduced. The improvement comes from collecting more diffracted information from the object.
319. A claim says, “A telescope with a larger objective is better only because it collects more light, not because of diffraction.” The better correction is that
ⓐ. a larger objective always increases \(\theta_{\min}\)
ⓑ. a larger objective reduces diffraction spreading
ⓒ. diffraction matters only in microscopes
ⓓ. objective diameter has no role in diffraction
Correct Answer: a larger objective reduces diffraction spreading
Explanation: A larger objective does collect more light, making images brighter. It also affects diffraction because the angular radius of the central diffraction disc is approximately \(\frac{1.22\lambda}{D_o}\). Increasing \(D_o\) decreases this angular radius. That reduces the minimum angular separation needed to resolve two point objects. The larger objective therefore improves both light-gathering ability and diffraction-limited resolving power.
320. A comparison of aperture effects is shown below.
| Case | Change made | Result |
|---|
| P | Single slit width \(a\) decreased | Diffraction spread increases |
| Q | Telescope objective diameter \(D_o\) increased | Minimum resolvable angle decreases |
| R | Single slit width \(a\) increased | Central maximum becomes narrower |
| S | Telescope objective diameter \(D_o\) increased | Minimum resolvable angle increases |
The row that should be corrected is
ⓐ. Row S
ⓑ. Row Q
ⓒ. Row P
ⓓ. Row R
Correct Answer: Row S
Explanation: Row P is correct because a narrower single slit gives more diffraction spreading. Row Q is correct because \(\theta_{\min}=\frac{1.22\lambda}{D_o}\), so increasing telescope aperture decreases the minimum resolvable angle. Row R is correct because a wider single slit produces a narrower central maximum. Row S is wrong because increasing \(D_o\) improves angular resolution rather than worsening it. Both slit diffraction and telescope resolution show that aperture size controls the spread of the diffraction pattern.
