Class 11 - Unit # 13 : Physical Optics - Solved Numericals

Physics Numerical Sheet
Physical Optics

1. A monochromatic light of wavelength $6900\text{ \AA}$ is used to illuminate two parallel slits. On a screen that is $3.30\text{ m}$ away from the slits, interference fringes are observed. The distance between adjacent bright fringes in the centre of the pattern is $1.80\text{ cm}$. What is the distance between the slits?

Data:

• Wavelength (\(\lambda\)) = \(6900\text{ \AA} = 6900 \times 10^{-10}\text{ m} = 6.9 \times 10^{-7}\text{ m}\)
• Distance to screen (\(L\)) = \(3.30\text{ m}\)
• Fringe spacing (\(\Delta y\)) = \(1.80\text{ cm} = 1.80 \times 10^{-2}\text{ m}\)
• Distance between slits (\(d\)) = ?

Solution:

In Young's Double Slit Experiment, the linear distance between two consecutive bright fringes (fringe width) is given by the relation:

$$\Delta y = \frac{\lambda L}{d}$$

Rearranging the formula to isolate the slit separation (\(d\)):

$$d = \frac{\lambda L}{\Delta y}$$ $$d = \frac{(6.9 \times 10^{-7}\text{ m}) \times 3.30\text{ m}}{1.80 \times 10^{-2}\text{ m}}$$ $$d = \frac{2.277 \times 10^{-6}}{1.80 \times 10^{-2}} = 1.265 \times 10^{-4}\text{ m} = 0.1265\text{ mm}$$

RESULT: The distance between the two slits is \(1.265 \times 10^{-4}\text{ m}\) (or \(0.1265\text{ mm}\)).

2. A Michelson interferometer is adjusted so that a bright fringe appears on the screen. As one of the mirrors is moved \(25.8\text{ }\mu\text{m}\), 92 bright fringes are counted crossing the field of view. What is the wavelength of light used in the interferometer?

Data:

• Distance shifted by the mirror (\(p\)) = \(25.8\text{ }\mu\text{m} = 25.8 \times 10^{-6}\text{ m}\)
• Number of fringes counted (\(m\)) = \(92\)
• Wavelength of light (\(\lambda\)) = ?

Solution:

In a Michelson Interferometer, moving a mirror by a distance $p$ changes the optical path length by $2p$. Each shift equal to half a wavelength (\(\lambda/2\)) causes exactly one fringe to migrate. The governing formula is:

$$2p = m\lambda$$

Isolating the wavelength (\(\lambda\)):

$$\lambda = \frac{2p}{m}$$ $$\lambda = \frac{2 \times (25.8 \times 10^{-6}\text{ m})}{92}$$ $$\lambda = \frac{51.6 \times 10^{-6}}{92} \approx 5.609 \times 10^{-7}\text{ m} = 560.9\text{ nm} = 5609\text{ \AA}$$

RESULT: The wavelength of the light source used is \(560.9\text{ nm}\) (or \(5609\text{ \AA}\)).

3. In study of optics, interference phenomena are observed due to thin films. Why must the film be thin? Why don't we see the interference effect when looking through a typical window or at a poster covered by a thick plate of glass, even if the glass is optically flat?

Answer:

To produce observable thin-film interference, the path difference between the light reflected from the upper surface and the lower surface must remain within the **coherence length** of the illuminating light source.

Two primary physical limitations prevent thick glass (like windows) from showing these patterns:

• Coherence Condition: Regular white light has a short coherence length. If a medium is thick (e.g., millimeters or centimeters), the path difference between the front and back reflections exceeds this length, making the waves mutually incoherent. They cannot create static phase relations.
• Fringe Overlap: Thick mediums pack fringes incredibly close together. Distinct spectral wavelengths produce maxima and minima that heavily overlap, blending back together into clean white light that masks any structural fringe systems.

Therefore, the medium thickness must be comparable to or a minor multiple of the incident light's wavelength (\(\sim\text{nanometers}\) to \(\mu\text{m}\)) to keep the paths coherent and the fringes visually resolvable.

4. Newton's rings are formed by light of \(400\text{ nm}\) wavelength. Determine the change in air film thickness between the third and sixth bright fringes. If the radius of curvature of the curved surface is \(5.0\text{ m}\), what is the radius of the third bright fringe?

Data:

• Wavelength (\(\lambda\)) = \(400\text{ nm} = 400 \times 10^{-9}\text{ m} = 4.0 \times 10^{-7}\text{ m}\)
• Radius of curvature (\(R\)) = \(5.0\text{ m}\)
• Order for lower bright ring (\(m_1\)) = 3
• Order for upper bright ring (\(m_2\)) = 6
• Change in thickness (\(\Delta t\)) = ?
• Radius of 3rd bright ring (\(r_3\)) = ?

Solution:

The operational condition for the construction of **bright** Newton's rings in reflected arrangements is given by:

$$2t = \left(m - \frac{1}{2}\right)\lambda \implies t = \left(m - \frac{1}{2}\right)\frac{\lambda}{2}$$

Part 1: Change in film thickness (\(\Delta t\))

$$t_3 = \left(3 - \frac{1}{2}\right)\frac{\lambda}{2} = 2.5\frac{\lambda}{2}$$ $$t_6 = \left(6 - \frac{1}{2}\right)\frac{\lambda}{2} = 5.5\frac{\lambda}{2}$$ $$\Delta t = t_6 - t_3 = (5.5 - 2.5)\frac{\lambda}{2} = 3\left(\frac{\lambda}{2}\right)$$ $$\Delta t = 3 \times \left(\frac{400 \times 10^{-9}\text{ m}}{2}\right) = 600 \times 10^{-9}\text{ m} = 600\text{ nm}$$

Part 2: Radius of the 3rd bright fringe (\(r_3\))

The mathematical geometric relation mapping ring radius to layer thickness is \(r^2 = 2tR\). Substituting the condition for the 3rd bright ring:

$$r_3^2 = 2 \left[\left(3 - \frac{1}{2}\right)\frac{\lambda}{2}\right] R = \left(2.5\right)\lambda R$$ $$r_3^2 = 2.5 \times (4.0 \times 10^{-7}\text{ m}) \times 5.0\text{ m} = 5.0 \times 10^{-6}\text{ m}^2$$ $$r_3 = \sqrt{5.0 \times 10^{-6}} \approx 2.236 \times 10^{-3}\text{ m} = 2.24\text{ mm}$$

RESULT: The change in air film thickness is \(600\text{ nm}\) and the radius of the third bright ring is \(2.24\text{ mm}\).

5. A soap film has an index of refraction \(n = 1.50\). The film is viewed in reflected light. (a) At a spot where the film thickness is \(910.0\text{ nm}\), which wavelengths are missing in the reflected light? (b) Which wavelengths are strongest in the visible light spectrum (400 nm - 700 nm)?

Data:

• Refractive index (\(n\)) = \(1.50\)
• Film thickness (\(t\)) = \(910.0\text{ nm}\)
• Constant term: \(2nt = 2 \times 1.50 \times 910.0\text{ nm} = 2730\text{ nm}\)

Solution:

Due to reflection at the boundary of an optically denser medium, a relative phase reversal of $\pi$ occurs at the upper air-to-film interface, introducing an intrinsic half-wavelength path penalty (\(\lambda / 2\)).

Part (a): Missing Wavelengths (Destructive Interference)

The net phase-inverted condition for destructive cancellation requires an integer multiple match:

$$2nt = m\lambda \implies \lambda = \frac{2nt}{m} = \frac{2730\text{ nm}}{m}$$

We solve systematically for integers ($m$) that yield wavelengths inside the visible range (\(400\text{ nm} - 700\text{ nm}\)):

• For \(m = 4\): \(\lambda = \frac{2730}{4} = 682.5\text{ nm}\)
• For \(m = 5\): \(\lambda = \frac{2730}{5} = 546.0\text{ nm}\)
• For \(m = 6\): \(\lambda = \frac{2730}{6} = 455.0\text{ nm}\)

Part (b): Strongest Wavelengths (Constructive Interference)

The path balance required to reinforce the light via constructive alignment is given by:

$$2nt = \left(m + \frac{1}{2}\right)\lambda \implies \lambda = \frac{2nt}{m + 0.5} = \frac{2730\text{ nm}}{m + 0.5}$$

Evaluating for integers ($m$) to collect values bounded inside visible limits:

• For \(m = 3\): \(\lambda = \frac{2730}{3.5} = 780.0\text{ nm}\) (Out of bounds - Infrared)
• For \(m = 4\): \(\lambda = \frac{2730}{4.5} = 606.7\text{ nm}\)
• For \(m = 5\): \(\lambda = \frac{2730}{5.5} = 496.4\text{ nm}\)
• For \(m = 6\): \(\lambda = \frac{2730}{6.5} = 420.0\text{ nm}\)

RESULT:
(a) Wavelengths missing (destructive): 682.5 nm, 546.0 nm, and 455.0 nm.
(b) Wavelengths strongest (constructive): 606.7 nm, 496.4 nm, and 420.0 nm.

12. The diffraction pattern from a single slit of width \(0.020\text{ mm}\) is viewed on a screen. If the screen is \(1.20\text{ m}\) from the slit and light of wavelength \(430\text{ nm}\) is used, what is the width of the central maximum?

Data:

• Slit width (\(a\)) = \(0.020\text{ mm} = 0.020 \times 10^{-3}\text{ m} = 2.0 \times 10^{-5}\text{ m}\)
• Screen distance (\(L\)) = \(1.20\text{ m}\)
• Wavelength (\(\lambda\)) = \(430\text{ nm} = 430 \times 10^{-9}\text{ m} = 4.3 \times 10^{-7}\text{ m}\)
• Linear width of central maximum (\(Y_c\)) = ?

Solution:

The angular position of the first diffraction minima framing the central core is given by \(a \sin\theta = \lambda\). Using the small-angle approximation (\(\sin\theta \approx \tan\theta = \frac{y}{L}\)):

$$y = \frac{\lambda L}{a}$$

The total legal span of the central maximum spreads symmetrically across both sides from center line to the first dark node (\(Y_c = 2y\)):

$$Y_c = \frac{2\lambda L}{a}$$ $$Y_c = \frac{2 \times (4.3 \times 10^{-7}\text{ m}) \times 1.20\text{ m}}{2.0 \times 10^{-5}\text{ m}}$$ $$Y_c = \frac{1.032 \times 10^{-6}}{2.0 \times 10^{-5}} = 0.0516\text{ m} = 5.16\text{ cm}$$

RESULT: The absolute spatial width of the central maximum is \(5.16\text{ cm}\).

13. A grating has exactly 8000 lines uniformly spaced over \(2.54\text{ cm}\) and is illuminated by light from a mercury vapor discharge lamp. What is the expected angle for the third-order maximum of light of wavelength \(546\text{ nm}\)?

Data:

• Total scale distance (\(w\)) = \(2.54\text{ cm} = 2.54 \times 10^{-2}\text{ m}\)
• Total number of lines (\(N\)) = \(8000\)
• Order of maximum (\(m\)) = 3
• Wavelength (\(\lambda\)) = \(546\text{ nm} = 546 \times 10^{-9}\text{ m}\)
• Diffraction angle (\(\theta\)) = ?

Solution:

First, calculate the grating element (the spatial distance between two adjacent lines, $d$):

$$d = \frac{w}{N} = \frac{2.54 \times 10^{-2}\text{ m}}{8000} = 3.175 \times 10^{-6}\text{ m}$$

Applying the standard diffraction grating equation:

$$d \sin\theta = m\lambda$$ $$\sin\theta = \frac{m\lambda}{d}$$ $$\sin\theta = \frac{3 \times (546 \times 10^{-9}\text{ m})}{3.175 \times 10^{-6}\text{ m}} = \frac{1.638 \times 10^{-6}}{3.175 \times 10^{-6}} \approx 0.5159$$

Calculating the principal angle value:

$$\theta = \sin^{-1}(0.5159) \approx 31.06^{\circ}$$

RESULT: The expected angular direction for the third-order maximum is \(31.06^{\circ}\).

14. How many lines per centimeter are there in a grating which gives a 1st-order spectrum at an angle of \(30^{\circ}\) when the wavelength of light is \(6 \times 10^{-5}\text{ cm}\)?

Data:

• Spectrum order (\(m\)) = 1
• Diffraction angle (\(\theta\)) = \(30^{\circ}\)
• Wavelength (\(\lambda\)) = \(6 \times 10^{-5}\text{ cm}\)
• Number of lines per centimeter (\(N\)) = ?

Solution:

The standard grating characteristic relationship is defined by:

$$d \sin\theta = m\lambda \implies d = \frac{m\lambda}{\sin\theta}$$

Substituting the values to find the grating element ($d$) in centimeters:

$$d = \frac{1 \times (6 \times 10^{-5}\text{ cm})}{\sin(30^{\circ})} = \frac{6 \times 10^{-5}}{0.5} = 1.2 \times 10^{-4}\text{ cm}$$

The total line count packed inside a unit length is the reciprocal of the grating element ($N = \frac{1}{d}$):

$$N = \frac{1}{1.2 \times 10^{-4}\text{ cm}} \approx 8333.33\text{ lines/cm}$$

RESULT: There are 8333 lines per centimeter configured on this grating.

15. Light of wavelength \(450\text{ nm}\) is incident on a diffraction grating on which 5000 lines/cm have been ruled. Determine (i) how many orders of spectra can be observed on either side of the normal and (ii) the angle corresponding to each order.

Data:

• Wavelength (\(\lambda\)) = \(450\text{ nm} = 450 \times 10^{-9}\text{ m} = 4.5 \times 10^{-7}\text{ m}\)
• Grating Density = \(5000\text{ lines/cm} = 500000\text{ lines/m}\)
• Grating element (\(d\)) = \(\frac{1}{5000}\text{ cm} = 2.0 \times 10^{-4}\text{ cm} = 2.0 \times 10^{-6}\text{ m} = 2000\text{ nm}\)

Solution:

Part (i): Maximum Observable Orders

The absolute physical upper bound for diffraction angles is \(\theta = 90^{\circ}\) (\(\sin\theta \le 1\)):

$$m_{\max} \le \frac{d \sin(90^{\circ})}{\lambda} = \frac{2000\text{ nm}}{450\text{ nm}} \approx 4.44$$

Since the order value must be an integer, the maximum observable order is **\(m = 4\)**.

Part (ii): Angular Direction calculations (\(\theta = \sin^{-1}(\frac{m\lambda}{d})\))

• Order 1 (\(m=1\)): \(\sin\theta_1 = \frac{1 \times 450}{2000} = 0.225 \implies \theta_1 = \sin^{-1}(0.225) \approx 13.00^{\circ}\)
• Order 2 (\(m=2\)): \(\sin\theta_2 = \frac{2 \times 450}{2000} = 0.450 \implies \theta_2 = \sin^{-1}(0.450) \approx 26.74^{\circ}\)
• Order 3 (\(m=3\)): \(\sin\theta_3 = \frac{3 \times 450}{2000} = 0.675 \implies \theta_3 = \sin^{-1}(0.675) \approx 42.45^{\circ}\)
• Order 4 (\(m=4\)): \(\sin\theta_4 = \frac{4 \times 450}{2000} = 0.900 \implies \theta_4 = \sin^{-1}(0.900) \approx 64.16^{\circ}\)

RESULT: A total of 4 spectral orders can be observed. The matching angles are \(13.00^{\circ}\), \(26.74^{\circ}\), \(42.45^{\circ}\), and \(64.16^{\circ}\).

16. Why does a crystal lattice act as an effective three-dimensional diffraction grating for X-rays but fails to do so for visible light waves?

Answer:

For diffraction to take place through any structured periodic array, the wavelength of the probing wave radiation ($\lambda$) must be less than or comparable to the spatial lattice spacing ($d$) separating the elements. This is explicitly stated in **Bragg's Law**:

$$2d \sin\theta = m\lambda \implies \sin\theta = \frac{m\lambda}{2d}$$

Because the trigonometric domain of sine values cannot physically exceed one (\(\sin\theta \le 1\)), the equation can only resolve if:

$$\lambda \le 2d$$

In typical crystals, the interplanar spacing ($d$) between atomic layers is roughly on the scale of fractions of a nanometer (\(\sim 0.1 - 0.3\text{ nm}\)).

• Visible Light possesses wavelengths spanning \(400 - 700\text{ nm}\), which are thousands of times larger than the crystal spacing (\(\lambda \gg 2d\)). This results in no real mathematical solution for $\theta$.
• X-rays naturally possess short wavelengths on the order of picometers to nanometers (\(\sim 0.01 - 1\text{ nm}\)), which perfectly match atomic dimensions.

Hence, crystals can only scatter and diffract X-rays effectively to form structural interference patterns.

17. A beam of X-rays of wavelength \(0.071\text{ nm}\) is diffracted by a diffracting plane of rock salt where the distance between the atomic planes is \(1.98\text{ \AA}\). Find the glancing angle for second-order diffraction.

Data:

• Wavelength (\(\lambda\)) = \(0.071\text{ nm} = 0.071 \times 10^{-9}\text{ m}\)
• Interplanar distance (\(d\)) = \(1.98\text{ \AA} = 1.98 \times 10^{-10}\text{ m} = 0.198 \times 10^{-9}\text{ m}\)
• Order of diffraction (\(m\)) = 2
• Glancing angle (\(\theta\)) = ?

Solution:

Applying Bragg's equation for crystal diffraction:

$$2d \sin\theta = m\lambda$$

Isolating the \(\sin\theta\) component:

$$\sin\theta = \frac{m\lambda}{2d}$$ $$\sin\theta = \frac{2 \times (0.071 \times 10^{-9}\text{ m})}{2 \times (0.198 \times 10^{-10}\text{ m})} = \frac{0.071}{0.198} \approx 0.3586$$

Calculating the inverse sine value:

$$\theta = \sin^{-1}(0.3586) \approx 21.01^{\circ}$$

RESULT: The glancing angle for second-order diffraction is \(21.01^{\circ}\).

18. Unpolarized light passes through two polarizers in turn with polarization axes configured at \(45^{\circ}\) to each other. What is the fraction of the initial incident light intensity that is successfully transmitted?

Data:

• Initial unpolarized light intensity = \(I_0\)
• Relative intersection angle (\(\theta\)) = \(45^{\circ}\)

Solution:

When completely unpolarized light passes through the initial linear polarizing filter, it loses half of its intensity because only parallel wave components pass through:

$$I_1 = \frac{I_0}{2}$$

When this newly polarized beam encounters the second filter, its transmission follows **Malus's Law**:

$$I_2 = I_1 \cos^2\theta$$ $$I_2 = \left(\frac{I_0}{2}\right) \cos^2(45^{\circ})$$

Since \(\cos(45^{\circ}) = \frac{1}{\sqrt{2}}\), its squared value is \(\frac{1}{2}\):

$$I_2 = \frac{I_0}{2} \times \left(\frac{1}{2}\right) = \frac{I_0}{4} = 0.25 I_0$$

Expressed as a percentage ratio:

$$\text{Fraction Transmitted} = \frac{I_2}{I_0} \times 100 = 25\%$$

RESULT: The total fraction of the incident light intensity transmitted through the filters is 25%.