Optics Numerical Solutions

phy-1109 PHY-1109 Physics

Topic: Optics (Interference, Diffraction, Polarization, Geometrical Optics)

Source: PYQs 2016–2024

	Subtopic	Question (Numerical Input/Calculation Required)	Year(s)
1	Young’s Double Slit	Young’s double slit experiment: slit separation 1.9 mm, fringe spacing 0.31 mm at 1 m distance. Calculate $\lambda$.	2016
2	Young’s Double Slit	Two coherent sources 0.22 mm apart. Screen 70 cm away. 4th bright fringe is 10 mm from central fringe. Calculate $\lambda$.	2017
3	Young’s Double Slit	Light of wavelength $5500\text{A˚}$ incident on double slit. Fringe separation $1$ cm on screen $200$ cm away. Calculate (i) slit separation and (ii) fringe width.	2020
4	Fresnel’s Bi-prism	Bi-prism calculation: $\lambda =5890\text{A˚}$, fringes 0.2 mm apart, $D=175$ cm. $\mu =1.5$. Prism is 5 cm from slit. Calculate angle at vertex.	2019
5	Fresnel’s Bi-prism	Bi-prism is placed 6 cm from slit. Fringes $0.05$ cm wide. $D=75$ cm. Find width of fringes observed if an eyepiece placed at 75 cm from the bi-prism.	2023
6	Newton’s Rings	Newton’s rings experiment: diameter of 5th ring was 0.336 cm, 15th ring=0.590 cm. Find the radius of curvature of the plano-convex lens, if the wavelength of the light used is $5890\text{A˚}$.	2016, 2022
7	Newton’s Rings	Newton’s ring calculation: diameter of 15th ring 0.59 cm, 5th ring 0.336 cm. $R=100$ cm. Calculate $\lambda$.	2018, 2024
8	Diffraction Grating	Deduce the missing order for a double slit Fraunhoffer diffraction pattern if the width of the slit and the opaque space are 0.16 and 0.8 mm respectively.	2018
9	Diffraction Grating	A plane grating has 15000 lines per inch. Find the angle of separation of the 5048 $\text{A˚}$ and 5016 $\text{A˚}$ lines of helium in the second order spectrum.	2016
10	Diffraction Grating	Plane transmission grating. Angle of diffraction for 2nd order principal maximum for $\lambda =5\times 10^{-5}$ cm is $30^{\circ }$. Calculate the number of lens (lines) in one cm of the grating surface.	2023
11	Polarization (Brewster)	Calculate the Brewster’s angle of light which travels from air into the glass medium ($\mu =1.55$) into the air medium.	2021
12	Polarization (Brewster)	Refractive index of crown glass 1.52. Find polarizing angle. If polarizing angle is about $59.50^{\circ }$, find refractive index of glass.	2023
13	Specific Rotation	Calculate the specific rotation of a given sample of sugar solution if the plane of polarization is turned through 26.4°. The length of the tube containing 20% sugar solution is 20 cm.	2017, 2019
14	Geometrical Optics	Two thin convex lenses $f_1=20$ cm, $f_2=2$ cm, separated by 2 cm. Calculate equivalent focal length and position of principal points.	2021

1. Young’s Double Slit Experiment

Q1: Calculate Wavelength

Year: 2016

Given:

• Slit separation ($d$) = $1.9\text{ mm}=1.9\times 10^{-3}\text{ m}$

• Fringe spacing/width ($\beta$) = $0.31\text{ mm}=0.31\times 10^{-3}\text{ m}$

• Distance to screen ($D$) = $1\text{ m}$

Formula:

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

Calculation:

\lambda = \frac{(0.31 \times 10^{-3})(1.9 \times 10^{-3})}{1}$$$$\lambda = 0.589 \times 10^{-6} \text{ m}

Answer: $5890\text{A˚}$ (or $589\text{ nm}$)

Q2: Calculate Wavelength from Fringe Position

Year: 2017

Given:

• $d=0.22\text{ mm}=0.22\times 10^{-3}\text{ m}$

• $D=70\text{ cm}=0.7\text{ m}$

• Position of 4th bright fringe ($x_4$) = $10\text{ mm}=10\times 10^{-3}\text{ m}$

• Order ($n$) = 4

Formula:

$x_n=\frac{n\lambda D}{d}\Rightarrow \lambda =\frac{x_{n}d}{nD}$

Calculation:

\lambda = \frac{(10 \times 10^{-3})(0.22 \times 10^{-3})}{4 \times 0.7}$$$$\lambda = \frac{2.2 \times 10^{-6}}{2.8} \approx 0.7857 \times 10^{-6} \text{ m}

Answer: $7857\text{A˚}$

Q3: Slit Separation and Fringe Width

Year: 2020

Given:

• $\lambda =5500\text{A˚}=5500\times 10^{-8}\text{ cm}$

• Fringe separation ($\beta$) = $1\text{ cm}$

• $D=200\text{ cm}$

Solution:

(i) Slit Separation ($d$):

d = \frac{\lambda D}{\beta} = \frac{5500 \times 10^{-8} \times 200}{1}$$$$d = 11 \times 10^{-4} \text{ cm} = 0.011 \text{ cm}

(ii) Fringe Width:

The question provides “Fringe separation 1 cm” and asks for “fringe width”. In YDSE, fringe separation is the fringe width.

Answer: Slit separation = 0.11 mm; Fringe width = 1 cm.

2. Fresnel’s Bi-prism

Q4: Calculate Angle at Vertex

Year: 2019

Given:

• $\lambda =5890\text{A˚}=5.89\times 10^{-5}\text{ cm}$

• $\beta =0.2\text{ mm}=0.02\text{ cm}$

• $D=175\text{ cm}$

• $\mu =1.5$

• Distance slit to prism ($u$) = $5\text{ cm}$

Formula:

For a biprism, the separation of virtual sources is $d=2u(\mu -1)\alpha$, where $\alpha$ is the refracting angle (in radians).

Substitute into $\beta =\frac{\lambda D}{d}$:

$\alpha =\frac{\lambda D}{2u(\mu -1)\beta }$

Calculation:

\alpha = \frac{5.89 \times 10^{-5} \times 175}{2 \times 5 \times (1.5 - 1) \times 0.02}$$$$\alpha = \frac{0.0103075}{10 \times 0.5 \times 0.02} = \frac{0.0103075}{0.1} = 0.103 \text{ rad}

Convert to degrees: $0.103\times \frac{180}{\pi }\approx 5.9^{\circ }$.

Answer: $0.103\text{ rad}$ (or approx $5.9^{\circ }$)

Q5: Change in Fringe Width

Year: 2023

Given:

• $u=6\text{ cm}$ (Slit to Prism)

• Initial ${\beta }_1=0.05\text{ cm}$ at $D_1=75\text{ cm}$ (assumed Slit to Screen)

• New condition: Eyepiece at 75 cm from bi-prism.

  • New $D_2=u+\text{distance(prism to eye)}=6+75=81\text{ cm}$.

Formula:

Since $\beta \propto D$:

$\frac{{\beta }_2}{{\beta }_1}=\frac{D_2}{D_1}$

Calculation:

${\beta }_2=0.05\times \frac{81}{75}=0.05\times 1.08$

Answer: $0.054\text{ cm}$

3. Newton’s Rings

Q6: Radius of Curvature

Year: 2016, 2022

Given:

• Diameter of 5th ring ($D_5$) = $0.336\text{ cm}$ ($m=5$)

• Diameter of 15th ring ($D_{15}$) = $0.590\text{ cm}$ ($n=15$)

• $\lambda =5890\text{A˚}=5.89\times 10^{-5}\text{ cm}$

Formula:

$R=\frac{D_n^2-D_m^2}{4(n-m)\lambda }$

Calculation:

R = \frac{(0.590)^2 - (0.336)^2}{4(15-5)(5.89 \times 10^{-5})}$$$$R = \frac{0.3481 - 0.112896}{40 \times 5.89 \times 10^{-5}} = \frac{0.235204}{0.002356}$$$$R \approx 99.83 \text{ cm}

Answer: $\approx 100\text{ cm}$

Q7: Calculate Wavelength

Year: 2018, 2024

Given: Same ring data as above, with $R=100\text{ cm}$.

Calculation:

Rearranging the formula from Q6:

\lambda = \frac{D_{15}^2 - D_5^2}{4(15-5)R} = \frac{0.2352}{40 \times 100}$$$$\lambda = 0.0000588 \text{ cm} = 5880 \times 10^{-8} \text{ cm}

Answer: $5880\text{A˚}$

4. Diffraction Grating

Q8: Missing Order (Absent Spectra)

Year: 2018

Given:

• Slit width ($a$) = $0.16\text{ mm}$

• Opaque width ($b$) = $0.8\text{ mm}$

Formula:

Condition for missing order:

$\frac{n}{\text{diffraction order}}=\frac{a+b}{a}$

Usually, we look for the order missing in the central diffraction maximum (diffraction order = 1).

Calculation:

$\text{Ratio}=\frac{0.16+0.8}{0.16}=\frac{0.96}{0.16}=6$

Answer: The 6th, 12th, 18th… orders of interference will be missing.

Q9: Angular Separation

Year: 2016

Given:

• $N=15000\text{ lines/inch}$

  • Grating element $(a+b)=\frac{2.54\text{ cm}}{15000}=1.6933\times 10^{-4}\text{ cm}$

• Order $n=2$

• ${\lambda }_1=5048\text{A˚}=5.048\times 10^{-5}\text{ cm}$

• ${\lambda }_2=5016\text{A˚}=5.016\times 10^{-5}\text{ cm}$

Formula:

$(a+b)\sin \theta =n\lambda \Rightarrow \theta ={\sin }^{-1}\left(\frac{n\lambda }{a+b}\right)$

Calculation:

For ${\lambda }_1$:

$\sin {\theta }_1=\frac{2\times 5.048\times 10^{-5}}{1.6933\times 10^{-4}}=0.5962\Rightarrow {\theta }_1\approx 36.60^{\circ }$

For ${\lambda }_2$:

$\sin {\theta }_2=\frac{2\times 5.016\times 10^{-5}}{1.6933\times 10^{-4}}=0.5924\Rightarrow {\theta }_2\approx 36.33^{\circ }$

Difference: $36.60^{\circ }-36.33^{\circ }=0.27^{\circ }$

Answer: 0.27° (approx)

Q10: Lines per cm

Year: 2023

Given:

• $n=2$

• $\lambda =5\times 10^{-5}\text{ cm}$

• $\theta =30^{\circ }$

Formula:

(a+b)\sin\theta = n\lambda$$$$(a+b) = \frac{1}{N} \quad (\text{where } N \text{ is lines/cm})$$$$\frac{1}{N}\sin(30^\circ) = 2 \times (5 \times 10^{-5})

Calculation:

\frac{1}{N} \times 0.5 = 10 \times 10^{-5}$$$$\frac{1}{N} = 20 \times 10^{-5} \Rightarrow N = \frac{1}{20 \times 10^{-5}} = \frac{10^5}{20} = 5000

Answer: $5000\text{ lines/cm}$

5. Polarization

Q11: Brewster’s Angle

Year: 2021

Given:

• $\mu =1.55$

Formula:

$\tan {\theta }_p=\mu$

Calculation:

${\theta }_p={\tan }^{-1}(1.55)$

Answer: $57.17^{\circ }$

Q12: Polarizing Angle and Refractive Index

Year: 2023

Part 1 Given: $\mu =1.52$

${\theta }_p={\tan }^{-1}(1.52)\approx 56.66^{\circ }$

Part 2 Given: ${\theta }_p=59.50^{\circ }$

$\mu =\tan (59.50^{\circ })\approx 1.70$

Answer: (i) $56.66^{\circ }$, (ii) $\mu =1.70$

Q13: Specific Rotation

Year: 2017, 2019

Given:

• Rotation angle $\theta =26.4^{\circ }$

• Concentration $C=20\%=0.2{\text{ g/cm}}^3$

• Length $l=20\text{ cm}=2\text{ dm}$ (Note: Length must be in decimeters for this formula)

Formula:

$S=\frac{\theta }{l\times C}$

Calculation:

$S=\frac{26.4}{2\times 0.2}=\frac{26.4}{0.4}$

Answer: $66^{\circ }{\text{ dm}}^{-1}{\text{ (g/cc)}}^{-1}$

6. Geometrical Optics

Q14: Equivalent Focal Length & Principal Points

Year: 2021

Given:

• $f_1=20\text{ cm}$, $f_2=2\text{ cm}$

• Separation $d=2\text{ cm}$

Formula (Equivalent Focal Length $F$):

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$$$\frac{1}{F} = \frac{1}{20} + \frac{1}{2} - \frac{2}{20 \times 2} = 0.05 + 0.5 - 0.05 = 0.5$$$$F = \frac{1}{0.5} = +2 \text{ cm}

Formula (Principal Points):

• Position of first principal point ($L_1{H}_1$): $\alpha =\frac{+Fd}{f_2}=\frac{2\times 2}{2}=+2\text{ cm}$ (2 cm to the right of the first lens).

• Position of second principal point ($L_2{H}_2$): $\beta =\frac{-Fd}{f_1}=\frac{-(2\times 2)}{20}=-0.2\text{ cm}$ (0.2 cm to the left of the second lens).

Answer: $F=2\text{ cm}$; Principal points at +2 cm from first lens and -0.2 cm from second lens.