Q.In a Young's double-slit experiment, the slits are separated by $0.28\ \text{mm}$ and the screen is placed $1.4\ \text{m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\ \text{cm}$. Determine the wavelength of light used in the experiment.

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.When monochromatic light of wavelength $\lambda$ is used in Young's double slit experiment, if I is the intensity of light at a point on the screen where path difference is $\frac{\lambda}{3}$, then the intensity of light at a point on the screen where path difference becomes $\lambda$ is (A) $2I$ (B) $3I$ (C) $4I$ (D) $I$

   ›Reveal solutionSolution

   The intensity in Young’s double-slit experiment depends on the phase difference via $I = I_0 \cos^2(\Delta\phi/2)$. For path difference $\lambda/3$, the phase is $2\pi/3$, giving $I = I_0/4$; for path difference $\lambda$, the phase is $2\pi$, giving $I_0$. Thus the second intensity is $4$ times the first, so the answer is $4I$.

   Concept & Intuition

   In Young’s double-slit experiment, the two waves from the slits are coherent and have the same amplitude. The intensity at any point on the screen is determined by the phase difference between them, which is directly proportional to the path difference. The key relation is:

   Phase difference $\Delta\phi = \frac{2\pi}{\lambda} \times \text{(path difference)}$

   Intensity $I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$, where $I_0$ is the maximum intensity (at a bright fringe).

   The problem gives two specific path differences; we compute the corresponding intensities and compare them.

   Step-by-step reasoning

   1. Find the phase difference for path difference $\lambda/3$

   $\Delta\phi_1 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$

   This is $120^\circ$, a point between a bright and a dark fringe.

   2. Compute the intensity $I$ at that point

   $I = I_0 \cos^2\left(\frac{\Delta\phi_1}{2}\right) = I_0 \cos^2\left(\frac{\pi}{3}\right) = I_0 \left(\frac{1}{2}\right)^2 = \frac{I_0}{4}$

   So $I = I_0/4$, meaning the given $I$ is one-quarter of the maximum possible intensity.

   3. Find the phase difference for path difference $\lambda$

   $\Delta\phi_2 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$

   This corresponds to a full cycle — a bright fringe (constructive interference). …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the rms value of the electric field of an electromagnetic wave propagating in free space is $30\sqrt{\pi}~\mathrm{V~m^{-1}}$, then the intensity of the wave is (A) $30~\mathrm{W~m^{-2}}$ (B) $3.75~\mathrm{W~m^{-2}}$ (C) $15~\mathrm{W~m^{-2}}$ (D) $7.5~\mathrm{W~m^{-2}}$

   ›Reveal solutionSolution

   The intensity of an electromagnetic wave in free space is given by $I = \frac{1}{2} c \varepsilon_0 E_0^2$, where $E_0$ is the peak electric field. Using the rms value $E_{\text{rms}} = 30\sqrt{\pi}~\mathrm{V/m}$, we find $I = 7.5~\mathrm{W/m^2}$, so the correct option is (D).

   The key concept here is the intensity of an electromagnetic wave — the average power per unit area it carries. For a wave in free space, intensity is directly related to the square of the electric field amplitude. Since the problem gives the rms (root-mean-square) field, we must first convert to the peak field, because the standard formula uses the amplitude.

   Why this approach works:

   The intensity $I$ is the time-averaged Poynting vector magnitude. For a sinusoidal wave, the average of $\sin^2$ over a cycle is $1/2$, so the intensity depends on the square of the peak field. The rms value is $E_{\text{rms}} = E_0 / \sqrt{2}$, so we can easily go from rms to peak and then compute intensity.

   Let’s work through it step by step.

   1. Recall the intensity formula for an EM wave in free space. The intensity (average power per unit area) is

   $I = \frac{1}{2} c \varepsilon_0 E_0^2,$

   where $c = 3 \times 10^8~\mathrm{m/s}$ is the speed of light, $\varepsilon_0 = 8.85 \times 10^{-12}~\mathrm{F/m}$ is the permittivity of free space, and $E_0$ is the peak electric field.

   2. Relate the given rms field to the peak field. For a sinusoidal wave,

   $E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \quad \Rightarrow \quad E_0 = \sqrt{2} \, E_{\text{rms}}.$

   Here $E_{\text{rms}} = 30\sqrt{\pi}~\mathrm{V/m}$, so

   $E_0 = \sqrt{2} \cdot 30\sqrt{\pi} = 30\sqrt{2\pi}~\mathrm{V/m}.$

   3. Substitute into the intensity formula.

   $I = \frac{1}{2} c \varepsilon_0 (30\sqrt{2\pi})^2.$

   Square the field:

   $(30\sqrt{2\pi})^2 = 900 \cdot 2\pi = 1800\pi.$

   So

   $I = \frac{1}{2} c \varepsilon_0 \cdot 1800\pi = 900\pi \, c \varepsilon_0.$

   4. Use the known value of $c \varepsilon_0$. A useful identity: $c \varepsilon_0 = \frac{1}{\mu_0 c}$, but more directly,

   $c \varepsilon_0 = \frac{1}{377}~\mathrm{S} \quad \text{(since the impedance of free space } \eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 377~\Omega\text{)}.$

   Actually, $c \varepsilon_0 = \frac{1}{\eta_0} \approx \frac{1}{377}~\mathrm{A/V}$. But let’s compute numerically:

   $c \varepsilon_0 = (3 \times 10^8) \times (8.85 \times 10^{-12}) = 2.655 \times 10^{-3}~\mathrm{A/V}.$

   Then

   $$$$

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3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.In Young’s double slit experiment, when a screen is placed at a distance of 1.2 m from the plane of the slits, the fringe width is 2 mm. If the setup is immersed in a liquid of refractive index 1.25 and the screen is moved to a distance of 1.8 m from the slits, then the fringe width is (A) 2.4 mm (B) 1.6 mm (C) 2.1 mm (D) 1.8 mm

   ›Reveal solutionSolution

   Fringe width depends on wavelength and geometry; immersion in a liquid reduces wavelength by refractive index, and moving the screen changes the distance. The new fringe width is 2.4 mm.

   The key idea in Young’s double slit experiment is that fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength of light in the medium, $D$ is the distance from the slits to the screen, and $d$ is the slit separation. When the setup is immersed in a liquid, the wavelength changes because the frequency stays the same but the speed of light decreases — so $\lambda_{\text{liquid}} = \frac{\lambda_{\text{air}}}{n}$, where $n$ is the refractive index. The slit separation $d$ is a physical constant and does not change.

   We are given two situations: first in air, then in a liquid with a different screen distance. The fringe width changes due to both the change in wavelength and the change in $D$. Let’s work through it step by step.

   1. Write the fringe width in air.

      In air, $\beta_1 = \frac{\lambda_{\text{air}} D_1}{d}$, with $D_1 = 1.2\ \text{m}$ and $\beta_1 = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$.

   2. Write the fringe width in the liquid.

      In the liquid, the wavelength becomes $\lambda_{\text{liquid}} = \frac{\lambda_{\text{air}}}{n}$, where $n = 1.25$. The new screen distance is $D_2 = 1.8\ \text{m}$. So

   $\beta_2 = \frac{\lambda_{\text{liquid}} D_2}{d} = \frac{\lambda_{\text{air}} D_2}{n d}.$

   3. Relate the two fringe widths. From the first expression, $\frac{\lambda_{\text{air}}}{d} = \frac{\beta_1}{D_1}$. Substitute this into the second: …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Two sound waves each of intensity $I$ are superimposed. If the phase difference between the waves is $\dfrac{\pi}{2}$, then the intensity of the resultant wave is (A) $2I$ (B) $3I$ (C) $4I$ (D) $I$

   ›Reveal solutionSolution

   When two waves of equal intensity $I$ interfere with a phase difference of $\pi/2$, the resultant intensity is simply $2I$ — the same as incoherent addition, because the cosine term in the interference formula vanishes.

   The key concept here is interference of coherent waves. When two waves of the same frequency and amplitude overlap, the resultant intensity depends on the phase difference $\phi$ between them. The formula for the resultant intensity when two waves of equal intensity $I_0$ interfere is:

   $I_{\text{resultant}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$

   Since both waves have intensity $I$, we have $I_1 = I_2 = I$. The phase difference is given as $\phi = \pi/2$.

   1. Plug in the values

   $I_{\text{resultant}} = I + I + 2\sqrt{I \cdot I} \cos\left(\frac{\pi}{2}\right)$

   2. Evaluate the cosine

   $\cos\left(\frac{\pi}{2}\right) = 0$

   3. Simplify The interference term vanishes, leaving: $I_{\text{resultant}} = I + I = 2I$ …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.In Young’s double slit experiment with light of wavelength $\lambda$, the intensity of light at a point on the screen where the path difference becomes $\dfrac{\lambda}{3}$ is (I is intensity of the central bright fringe) (A) I (B) $\dfrac{I}{2}$ (C) $\dfrac{I}{3}$ (D) $\dfrac{I}{4}$

   ›Reveal solutionSolution

   The intensity at a point with path difference $\lambda/3$ is $I/4$, because the phase difference is $2\pi/3$ and the two-slit interference formula gives $I_\text{net} = 4I_0 \cos^2(\Delta\phi/2)$, with $I = 4I_0$ at the central maximum.

   The key idea is that in Young’s double‑slit experiment, the intensity at any point on the screen depends only on the phase difference between the two waves arriving there. The central bright fringe (path difference = 0) has maximum intensity $I$. When the path difference is $\lambda/3$, the phase difference is $2\pi/3$, and the interference formula directly gives the reduced intensity.

   ---

   Why this approach works

   The superposition of two identical waves (same amplitude $A_0$, same frequency) yields a resultant amplitude $A = 2A_0 \cos(\Delta\phi/2)$, where $\Delta\phi$ is the phase difference. Intensity is proportional to the square of amplitude, so $I_\text{net} = 4I_0 \cos^2(\Delta\phi/2)$, where $I_0$ is the intensity from one slit alone. At the central maximum, $\Delta\phi = 0$, so $I = 4I_0$. Once we find $\Delta\phi$ from the given path difference, we can compute the ratio $I_\text{net}/I$.

   ---

   Step‑by‑step solution

   1. Relate path difference to phase difference. For a wave of wavelength $\lambda$, a path difference $\Delta x$ corresponds to a phase difference

   $\Delta\phi = \frac{2\pi}{\lambda} \cdot \Delta x.$

   Here $\Delta x = \lambda/3$, so

   $\Delta\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}.$

   2. Write the general intensity formula for two‑slit interference. If each slit alone produces intensity $I_0$ at the screen, the combined intensity when the phase difference is $\Delta\phi$ is

   $I_{\text{net}} = 4I_0 \cos^2\!\left(\frac{\Delta\phi}{2}\right).$

   This comes from $A = 2A_0 \cos(\Delta\phi/2)$ and $I \propto A^2$.

   3. Find $I_0$ in terms of the central maximum intensity $I$. At the central bright fringe, $\Delta\phi = 0$, so $\cos(0) = 1$ and $I = 4I_0 \quad\Rightarrow\quad I_0 = \frac{I}{4}.$ …

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6. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.The wavelength of the monochromatic light used in Young's double slit experiment is 550 nm and the screen is placed at a distance of 120 cm from the plane of the slits. If third dark fringe is formed on the screen at a distance of 1.5 mm from the central bright fringe, then the distance of separation between the two slits is (A) 5.5 mm (B) 1.1 mm (C) 2.2 mm (D) 3.3 mm

   ›Reveal solutionSolution

   This problem involves Young's Double Slit Experiment, specifically the position of a dark fringe. We use the formula for the position of the $n$-th dark fringe to find the slit separation. The distance of separation between the two slits is $\boxed{\text{1.1 mm}}$.

   In Young's Double Slit Experiment (YDSE), when monochromatic light passes through two narrow slits, an interference pattern of alternating bright and dark fringes is observed on a screen. This pattern arises from the superposition of waves from the two slits.

   Concept of Dark Fringes

   Dark fringes occur at points on the screen where the waves from the two slits interfere destructively. Destructive interference happens when the path difference between the waves arriving at a point is an odd multiple of half the wavelength.

   That is, path difference $\Delta x = (n - \frac{1}{2})\lambda$, where $n = 1, 2, 3, \dots$ for the first, second, third dark fringes, and so on.

   For a typical YDSE setup where the screen is far from the slits ($D \gg d$), the position $y_n$ of the $n$-th dark fringe from the central bright fringe is given by:

   $y_n = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d}$

   where:

   • $y_n$ is the distance of the $n$-th dark fringe from the central bright fringe.
   • $n$ is the order of the dark fringe ($n=1$ for the first, $n=2$ for the second, etc.).
   • $\lambda$ is the wavelength of the light.
   • $D$ is the distance between the slits and the screen.
   • $d$ is the distance of separation between the two slits.

   We are given the values for $\lambda$, $D$, $y_n$, and the order of the dark fringe ($n=3$). We need to find $d$.

   1. Identify Given Parameters and Convert to SI Units:

      • Wavelength of light, $\lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m}$
      • Distance of the screen from the slits, $D = 120 \text{ cm} = 1.20 \text{ m}$
      • Position of the third dark fringe, $y_3 = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}$
      • Order of the dark fringe, $n = 3$

   2. Apply the Formula for the Position of the Dark Fringe:

      For the third dark fringe, we set $n=3$ in the formula:

      $y_3 = \left(3 - \frac{1}{2}\right) \frac{\lambda D}{d}$ …

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7. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.In an experiment, the angular width of interference fringes for a light of wavelength 5896 $\text{\AA}$ is found to be $3.5 \times 10^{-3}$ radian. The wavelength of light for which the angular width of the fringes becomes 10% greater is (A) 5306.4 $\text{\AA}$ (B) 5886 $\text{\AA}$ (C) 5906 $\text{\AA}$ (D) 6485.6 $\text{\AA}$

   ›Reveal solutionSolution

   The angular width of interference fringes is directly proportional to the wavelength of light. To achieve a 10% greater angular width, the wavelength must also increase by 10%, leading to a new wavelength of $\boxed{6485.6 \text{ Å}}$.

   The phenomenon of interference produces a pattern of alternating bright and dark fringes on a screen. The angular width of these fringes refers to the angle subtended by the distance between two consecutive bright or dark fringes at the position of the slits. This angular width is a crucial characteristic of the interference pattern, as it describes how spread out the fringes appear.

   For a Young's Double Slit Experiment (YDSE), the linear fringe width, $\beta$, which is the distance between two consecutive bright or dark fringes on the screen, is given by the formula:

   $\beta = \frac{\lambda D}{d}$

   where $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the two slits.

   The angular width, $\theta$, is defined as the linear fringe width divided by the distance to the screen:

   $\theta = \frac{\beta}{D}$

   Substituting the expression for $\beta$:

   $\theta = \frac{(\lambda D / d)}{D} = \frac{\lambda}{d}$

   The angular width of interference fringes is given by:

   $\theta = \frac{\lambda}{d}$

   where $\lambda$ is the wavelength of light and $d$ is the distance between the slits.

   From this formula, it is clear that for a given experimental setup (where $d$ is constant), the angular width $\theta$ is directly proportional to the wavelength $\lambda$ of the light used.

   $\theta \propto \lambda$

   This direct proportionality is the key concept for solving this problem. If the angular width increases, the wavelength must also increase by the same proportion, assuming the slit separation $d$ remains unchanged.

   Let's apply this understanding to the problem:

   1. Identify the initial conditions:

      We are given the initial wavelength of light, $\lambda_1$, and the initial angular width of the fringes, $\theta_1$.

      • Initial wavelength, $\lambda_1 = 5896 \text{ Å}$
      • Initial angular width, $\theta_1 = 3.5 \times 10^{-3} \text{ radian}$

   2. Determine the new angular width:

      The problem states that the angular width of the fringes becomes 10% greater. This means the new angular width, $\theta_2$, will be $100\% + 10\% = 110\%$ of the initial angular width.

      • $\theta_2 = \theta_1 + 0.10 \theta_1 = 1.10 \theta_1$
      • $\theta_2 = 1.10 \times (3.5 \times 10^{-3} \text{ radian})$
      • $\theta_2 = 3.85 \times 10^{-3} \text{ radian}$ …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.Young’s double slit experiment is performed with monochromatic light of wavelength 6000 $\text{\AA}$. If the intensity of light at a point on the screen where path difference of 2000 $\text{\AA}$ is $I_1$ and the intensity of light at a point on the screen where the path difference is 1000 $\text{\AA}$ is $I_2$, then $I_1 : I_2 =$ (A) $1 : 3$ (B) $2 : 1$ (C) $1 : 1$ (D) $4 : 5$

   ›Reveal solutionSolution

   In Young’s double slit experiment, intensity depends on the phase difference via $I = I_0 \cos^2(\Delta\phi/2)$. The path differences given correspond to phase differences that yield intensity ratio $I_1 : I_2 = 1 : 3$, so the correct option is (A).

   Concept and intuition

   In Young’s double slit experiment, the intensity at any point on the screen is not simply proportional to the path difference itself, but to the cosine-squared of half the phase difference. The phase difference is $\Delta\phi = \frac{2\pi}{\lambda} \times (\text{path difference})$. So the key is: convert each path difference into a phase difference, then compute the relative intensities. A common mistake is to think intensity is proportional to path difference directly — that would give a wrong ratio. Instead, the interference pattern’s brightness oscillates sinusoidally.

   Step-by-step solution

   1. Write the relation between path difference and phase difference For a wavelength $\lambda$, a path difference $\Delta x$ gives a phase difference

   $\Delta\phi = \frac{2\pi}{\lambda} \Delta x.$

   Here $\lambda = 6000\ \text{Å}$.

   2. Compute the phase difference for each case
      • For $\Delta x_1 = 2000\ \text{Å}$:

   $\Delta\phi_1 = \frac{2\pi}{6000} \times 2000 = \frac{2\pi}{3}.$

   • For $\Delta x_2 = 1000\ \text{Å}$:

   $\Delta\phi_2 = \frac{2\pi}{6000} \times 1000 = \frac{\pi}{3}.$

   3. Recall the intensity formula In Young’s double slit, if each slit alone gives intensity $I_0$, the combined intensity at a phase difference $\Delta\phi$ is

   $I = 4I_0 \cos^2\left(\frac{\Delta\phi}{2}\right).$

   (The factor $4I_0$ is the maximum intensity; it cancels in a ratio.)

   4. Find the intensities $I_1$ and $I_2$
      • For $\Delta\phi_1 = 2\pi/3$: $$I_1 = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \cdot \frac{1}{4} = I_0. …$$

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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The work done in blowing a soap bubble of volume $V$ is $W$. The work done in blowing the bubble of volume $2V$ from the same soap solution is (A) $\dfrac{W}{2}$ (B) $\sqrt{2} W$ (C) $(2)^{\frac{1}{3}} W$ (D) $(4)^{\frac{1}{3}} W$

   ›Reveal solutionSolution

   The work done in blowing a soap bubble is proportional to its surface area, not its volume. Since surface area scales as (volume)^{2/3}, doubling the volume multiplies the work by $2^{2/3} = 4^{1/3}$. The correct option is (D).

   The key concept here is that work done in blowing a soap bubble equals the increase in surface energy, which is proportional to the surface area of the bubble (since the soap solution has constant surface tension). A common mistake is to think work scales with volume — but a bubble has two surfaces (inner and outer), and the work is entirely about stretching the film.

   1. Relate work to surface area For a soap bubble of radius $r$, the total surface area (both inner and outer surfaces) is $2 \times 4\pi r^2 = 8\pi r^2$. The work done $W$ is the product of surface tension $T$ and the increase in area:

   $W = T \cdot 8\pi r^2.$

   So $W \propto r^2$.

   2. Relate radius to volume Volume of a sphere is $V = \frac{4}{3}\pi r^3$, so $r \propto V^{1/3}$. Therefore, surface area $\propto r^2 \propto V^{2/3}$, and hence

   $W \propto V^{2/3}.$

   3. Apply to the new volume If the original volume is $V$, work is $W = k V^{2/3}$ for some constant $k$. For volume $2V$, the work $W'$ is $W' = k (2V)^{2/3} = k V^{2/3} \cdot 2^{2/3} = W \cdot 2^{2/3}.$ …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.An inductor of inductive reactance $R$, a capacitor of capacitive reactance $2R$ and a resistor of resistance $R$ are connected in series to an ac source. The power factor of the series LCR circuit is (A) $\dfrac{1}{\sqrt{2}}$ (B) $\dfrac{1}{\sqrt{3}}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{\sqrt{3}}{2}$

    ›Reveal solutionSolution

    The power factor is the cosine of the phase angle between voltage and current, given by $R/Z$. Here $X_L = R$, $X_C = 2R$, so net reactance $X = -R$, impedance $Z = \sqrt{R^2 + R^2} = R\sqrt{2}$, giving power factor $1/\sqrt{2}$. The correct option is (A).

    Concept & Intuition

    In an AC series circuit, the power factor tells us what fraction of the apparent power is actually doing real work. It equals $\cos\phi$, where $\phi$ is the phase difference between voltage and current. For a series LCR circuit, the impedance $Z$ combines resistance $R$ and net reactance $X = X_L - X_C$ via $Z = \sqrt{R^2 + X^2}$. The power factor is then $\cos\phi = R/Z$. The trick here is that the inductive reactance and capacitive reactance are given as multiples of $R$, so we can compute $X$ and $Z$ purely in terms of $R$.

    Step-by-step solution

    1. Identify the given reactances

       Inductive reactance $X_L = R$ (yes, the problem uses $R$ for both resistance and reactance — careful!).

       Capacitive reactance $X_C = 2R$.

       Resistance $R$ (same symbol, but it’s the resistor’s value).

    2. Find the net reactance

       In a series circuit, net reactance $X = X_L - X_C = R - 2R = -R$.

       The negative sign means the circuit is capacitive overall (current leads voltage), but magnitude is what matters for impedance.

    3. Compute the impedance magnitude

       $Z = \sqrt{R^2 + X^2} = \sqrt{R^2 + (-R)^2} = \sqrt{2R^2} = R\sqrt{2}.$ …

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.If the slit width is 2 mm and wavelength of light used is 4000 $\text{\AA}$, then Fresnel distance is nearly (A) 2 mm (B) 10 m (C) 20 km (D) 2 $\mu$m

    ›Reveal solutionSolution

    The Fresnel distance is the distance at which diffraction effects become significant; it is given by $a^2/\lambda$. With slit width $a = 2\ \text{mm} = 2\times10^{-3}\ \text{m}$ and wavelength $\lambda = 4000\ \text{Å} = 4\times10^{-7}\ \text{m}$, the Fresnel distance is $10\ \text{m}$, so option (B) is correct.

    Concept and intuition:

    The Fresnel distance (or Rayleigh distance) marks the boundary between near-field (Fresnel) and far-field (Fraunhofer) diffraction. For a slit of width $a$, when the observation distance $D$ is much less than $a^2/\lambda$, the wavefront curvature matters (Fresnel regime). When $D \gg a^2/\lambda$, the wavefront is essentially planar (Fraunhofer regime). The quantity $a^2/\lambda$ is the characteristic distance where the phase difference across the slit becomes about 1 radian. So we simply compute that value.

    Step-by-step solution:

    1. Identify given quantities

       Slit width: $a = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$

       Wavelength: $\lambda = 4000\ \text{Å} = 4000 \times 10^{-10}\ \text{m} = 4 \times 10^{-7}\ \text{m}$

    2. Recall the Fresnel distance formula

       The Fresnel distance $D_F$ is defined as:

    $D_F = \frac{a^2}{\lambda}$

    3. Plug in the numbers

    $D_F = \frac{(2 \times 10^{-3})^2}{4 \times 10^{-7}} = \frac{4 \times 10^{-6}}{4 \times 10^{-7}} = 10\ \text{m}$

    4. Interpret the result …

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12. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.A block of mass ‘m’ is attached to the lower end of a spring whose upper end is fixed to a rigid support. The spring has negligible mass. When the block is slightly pulled down and released, it oscillates with a time period of 3 seconds. When the mass of the block is increased by 1 kg, the time period of oscillation becomes 5 seconds, then the value of ‘m’ in kg is (A) $\dfrac{3}{4}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{16}{9}$ (D) $\dfrac{9}{16}$

    ›Reveal solutionSolution

    The time period of a spring-mass system scales as $\sqrt{m}$. Writing two equations for the two given periods and taking their ratio eliminates the spring constant, leaving a simple algebraic relation between the masses that yields $m = \dfrac{9}{16}$ kg.

    The time period of a mass oscillating on a spring depends on both the mass and the stiffness of the spring. The fundamental relationship is

    $T = 2\pi\sqrt{\frac{m}{k}}$

    where $k$ is the spring constant. Notice that $T \propto \sqrt{m}$ when $k$ is fixed. This square-root dependence is the key: doubling the mass does not double the period; it increases it by a factor of $\sqrt{2}$.

    When we change the mass but keep the same spring, we can set up two equations and eliminate the unknown spring constant by taking a ratio. That ratio will contain only the masses and periods, both of which we know (or are solving for).

    ---

    Setting up the two scenarios:

    1. Original mass $m$, period $T_1 = 3$ s:

    $3 = 2\pi\sqrt{\frac{m}{k}}$

    2. Increased mass $m + 1$ kg, period $T_2 = 5$ s:

    $5 = 2\pi\sqrt{\frac{m+1}{k}}$

    Taking the ratio to eliminate $k$ and $2\pi$:

    3. Divide the second equation by the first:

       $\frac{5}{3} = \frac{2\pi\sqrt{\frac{m+1}{k}}}{2\pi\sqrt{\frac{m}{k}}} = \sqrt{\frac{m+1}{m}}$ …

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