💡 Chapter 14: Light – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 14 Light including refractive index, Snell's law, critical angle, mirror and lens formula, magnification, and power of lens. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 14: Light
Light covers reflection, refraction, total internal reflection, optical fibres, lenses, prisms, dispersion, and optical instruments. Includes solved examples and numerical problems.
📐 Numerical Problems & Solutions (PECTAA 2026)
14.1 Refractive Index of Water
The speed of light in water is \(2.25 \times 10^8 \, \text{ms}^{-1}\). Calculate the refractive index of water.
Given Data
\(Speed\ of\ light\ in\ air\)\(= c = 3 \times 10^8 \, \text{ms}^{-1}\)
\(Speed\ of\ light\ in\ water\)\(= v = 2.25 \times 10^8 \, \text{ms}^{-1}\)
To Find
\(Refractive\ index\ of\ water = n = ?\)
Solution
By using formula of refractive index:
\[\begin{aligned}
n &= \frac{c}{v} \\
n &= \frac{3 \times 10^8}{2.25 \times 10^8} \\
\boldsymbol{n} &= \boldsymbol{1.33}
\end{aligned}\]
✅ \(n = 1.33\)
14.2 Angle of Refraction (Snell's Law)
A light ray in air (\(n_1 = 1.0\)) strikes a glass surface (\(n_2 = 1.5\)) at an angle of incidence \(30^\circ\). Find the angle of refraction inside the glass.
Given Data
\(Refractive\ index\ of\ air\)\(= n_1 = 1.0\)
\(Refractive\ index\ of\ glass\)\(= n_2 = 1.5\)
\(Angle\ of\ incidence\)\(= i = 30^\circ\)
To Find
\(Angle\ of\ refraction = r = ?\)
Solution
By using Snell's law:
\[\begin{aligned}
n_1 \sin i &= n_2 \sin r \\
(1.0) \sin 30^\circ &= (1.5) \sin r \\
(1.0)(0.5) &= (1.5) \sin r \\
0.5 &= (1.5) \sin r \\
\sin r &= \frac{0.5}{1.5} = \frac{1}{3} \\
r &= \sin^{-1}\left(\frac{1}{3}\right) \\
\boldsymbol{r} &= \boldsymbol{19.5^\circ}
\end{aligned}\]
✅ \(r = 19.5^\circ\)
14.3 Refraction Through Glass Slab
A beam of light in air hits a glass slab at an incidence angle of \(40^\circ\). Find the angle of refraction inside the glass.
Given Data
\(Angle\ of\ incidence\)\(= i = 40^\circ\)
\(Refractive\ index\ of\ air\)\(= n_1 = 1.0\)
\(Refractive\ index\ of\ glass\)\(= n_2 = 1.5\)
To Find
\(Angle\ of\ refraction = r = ?\)
Solution
By using Snell's law:
\[\begin{aligned}
n_1 \sin i &= n_2 \sin r \\
(1.0) \sin 40^\circ &= (1.5) \sin r \\
(1.0)(0.6428) &= (1.5) \sin r \\
0.6428 &= (1.5) \sin r \\
\sin r &= \frac{0.6428}{1.5} = 0.4285 \\
r &= \sin^{-1}(0.4285) \\
\boldsymbol{r} &= \boldsymbol{25.4^\circ}
\end{aligned}\]
✅ \(r = 25.4^\circ\)
14.4 Refractive Index of Unknown Liquid
A ray of light travels from air into an unknown liquid. The angle of incidence in air is \(45^\circ\), and the angle of refraction in the liquid is \(30^\circ\). Calculate the refractive index of the liquid.
Given Data
\(Angle\ of\ incidence\)\(= i = 45^\circ\)
\(Angle\ of\ refraction\)\(= r = 30^\circ\)
To Find
\(Refractive\ index\ of\ liquid = n = ?\)
Solution
By using Snell's law:
\[\begin{aligned}
n &= \frac{\sin i}{\sin r} \\
n &= \frac{\sin 45^\circ}{\sin 30^\circ} \\
n &= \frac{0.7071}{0.5} \\
\boldsymbol{n} &= \boldsymbol{1.41}
\end{aligned}\]
✅ \(n = 1.41\)
14.5 Refractive Index and Speed of Light in Liquid
A ray of light passes from air into a liquid. The angle of incidence in air is \(50^\circ\), and the angle of refraction inside the liquid is \(32^\circ\). Calculate: (a) the refractive index of the liquid (b) the speed of light in the liquid.
Given Data
\(Angle\ of\ incidence\)\(= i = 50^\circ\)
\(Angle\ of\ refraction\)\(= r = 32^\circ\)
\(Speed\ of\ light\ in\ air\)\(= c = 3 \times 10^8 \, \text{ms}^{-1}\)
To Find
\(Refractive\ index\ of\ liquid = n = ?\)
\(Speed\ of\ light\ in\ liquid = v = ?\)
Solution
By using Snell's law:
\[\begin{aligned}
n &= \frac{\sin i}{\sin r} \\
n &= \frac{\sin 50^\circ}{\sin 32^\circ} \\
n &= \frac{0.7660}{0.5299} = 1.45
\end{aligned}\]
By using formula of refractive index:
\[\begin{aligned}
n &= \frac{c}{v} \\
v &= \frac{c}{n} \\
v &= \frac{3 \times 10^8}{1.45} \\
\boldsymbol{v} &= \boldsymbol{2.07 \times 10^8 \, \text{ms}^{-1}}
\end{aligned}\]
✅ \(n = 1.45\) | \(v = 2.07 \times 10^8 \, \text{ms}^{-1}\)
14.6 Critical Angle for Total Internal Reflection
Light travels inside a glass block of refractive index \(n = 1.6\). It strikes the glass-air boundary. Calculate the critical angle for total internal reflection.
Given Data
\(Refractive\ index\ of\ glass\)\(= n = 1.6\)
To Find
\(Critical\ angle = C = ?\)
Solution
By using formula of critical angle:
\[\begin{aligned}
C &= \sin^{-1}\left(\frac{1}{n}\right) \\
C &= \sin^{-1}\left(\frac{1}{1.6}\right) \\
\boldsymbol{C} &= \boldsymbol{38.7^\circ}
\end{aligned}\]
✅ \(C = 38.7^\circ\)
14.7 Focal Length of Convex Mirror
An object is placed 15.0 cm in front of a convex mirror, forming a virtual image 7.5 cm behind the mirror. What is the focal length of the mirror?
Given Data
\(Object\ distance\)\(= p = 15 \, \text{cm}\)
\(Image\ distance\)\(= q = -7.5 \, \text{cm (virtual, behind mirror)}\)
To Find
\(Focal\ length = f = ?\)
Solution
By using mirror formula:
\[\begin{aligned}
\frac{1}{f} &= \frac{1}{p} + \frac{1}{q} \\
\frac{1}{f} &= \frac{1}{15} + \frac{1}{-7.5} \\
\frac{1}{f} &= \frac{1}{15} - \frac{1}{7.5} \\
\frac{1}{f} &= \frac{7.5 - 15}{(15)(7.5)} \\
\frac{1}{f} &= \frac{-7.5}{112.5} \\
\boldsymbol{f} &= \boldsymbol{-15 \, \text{cm}}
\end{aligned}\]
✅ \(f = -15 \, \text{cm}\)
14.8 Image Location and Height (Concave Mirror)
An object 30 cm tall is located 10.5 cm from a concave mirror with focal length 16 cm. (a) Where is the image located? (b) How high is it?
Given Data
\(Object\ height\)\(= h_o = 30 \, \text{cm}\)
\(Object\ distance\)\(= p = 10.5 \, \text{cm}\)
\(Focal\ length\)\(= f = 16 \, \text{cm}\)
To Find
\(Image\ distance = q = ?\)
\(Image\ height = h_i = ?\)
Solution
By using mirror formula:
\[\begin{aligned}
\frac{1}{f} &= \frac{1}{p} + \frac{1}{q} \\
\frac{1}{q} &= \frac{1}{f} - \frac{1}{p} \\
\frac{1}{q} &= \frac{1}{16} - \frac{1}{10.5} \\
\frac{1}{q} &= \frac{10.5 - 16}{(16)(10.5)} \\
\frac{1}{q} &= \frac{-5.5}{168} \\
q &= -\frac{168}{5.5} = -30.54 \, \text{cm}
\end{aligned}\]
Negative sign shows that the image is virtual and located behind the mirror.
Now by using formula of magnification:
\[\begin{aligned}
M &= \frac{h_i}{h_o} = \frac{|q|}{p} \\
h_i &= \frac{|q|}{p} \times h_o \\
h_i &= \frac{30.54}{10.5} \times 30 \\
\boldsymbol{h_i} &= \boldsymbol{87.3 \, \text{cm}}
\end{aligned}\]
✅ \(q = -30.5 \, \text{cm}\) | \(h_i = 87.3 \, \text{cm}\)
14.9 Image Distance for Concave Lens
An object is placed at 30 cm to the left of a concave lens with focal length \(-15\) cm. Find the image distance and describe the image.
Given Data
\(Object\ distance\)\(= p = 30 \, \text{cm}\)
\(Focal\ length\)\(= f = -15 \, \text{cm (concave lens)}\)
To Find
\(Image\ distance = q = ?\)
\(Nature\ of\ image = ?\)
Solution
By using lens formula:
\[\begin{aligned}
\frac{1}{f} &= \frac{1}{p} + \frac{1}{q} \\
\frac{1}{q} &= \frac{1}{f} - \frac{1}{p} \\
\frac{1}{q} &= \frac{1}{-15} - \frac{1}{30} \\
\frac{1}{q} &= \frac{-1}{15} - \frac{1}{30} \\
\frac{1}{q} &= \frac{-2 - 1}{30} = \frac{-3}{30} \\
q &= -\frac{30}{3} = -10 \, \text{cm}
\end{aligned}\]
Since \(q\) is negative, the image is virtual, upright, and formed at a distance of 10 cm on the same side of the object.
Note: The textbook solution key mentions \(-30\) cm as a typo. The mathematically precise answer is \(-10\) cm.
✅ \(q = -10 \, \text{cm}\) (Virtual, Upright)
📘 Examples with Solutions
Ex 14.1 Refraction from Air to Water
A ray of light enters from air into water. The angle of incidence is \(40^\circ\) and the refractive index of water is 1.33. Find the angle of refraction.
Given Data
\(Angle\ of\ incidence\)\(= i = 40^\circ\)
\(Refractive\ index\ of\ water\)\(= n = 1.33\)
To Find
\(Angle\ of\ refraction = r = ?\)
Solution
By using Snell's law:
\[\begin{aligned}
\frac{\sin i}{\sin r} &= n \\
\sin r &= \frac{\sin i}{n} \\
\sin r &= \frac{\sin 40^\circ}{1.33} \\
\sin r &= \frac{0.6428}{1.33} = 0.4833 \\
r &= \sin^{-1}(0.4833) \\
\boldsymbol{r} &= \boldsymbol{28.9^\circ}
\end{aligned}\]
✅ \(r = 28.9^\circ\)
Ex 14.2 Speed of Light in Water
The refractive index of water is 1.33. Find the speed of light in water.
Given Data
\(Refractive\ index\ of\ water\)\(= n = 1.33\)
\(Speed\ of\ light\ in\ air\)\(= c = 3 \times 10^8 \, \text{ms}^{-1}\)
To Find
\(Speed\ of\ light\ in\ water = v = ?\)
Solution
By using formula of refractive index:
\[\begin{aligned}
n &= \frac{c}{v} \\
v &= \frac{c}{n} \\
v &= \frac{3 \times 10^8}{1.33} \\
\boldsymbol{v} &= \boldsymbol{2.26 \times 10^8 \, \text{ms}^{-1}}
\end{aligned}\]
✅ \(v = 2.26 \times 10^8 \, \text{ms}^{-1}\)
Ex 14.3 Critical Angle for Water
Find the value of critical angle for water. The refractive index of water is 1.33 and that of air is 1.0.
Given Data
\(Refractive\ index\ of\ water\)\(= n_1 = 1.33\)
\(Refractive\ index\ of\ air\)\(= n_2 = 1.0\)
\(Refracted\ angle\)\(= r = 90^\circ\)
To Find
\(Critical\ angle = C = ?\)
Solution
By using Snell's law (for critical angle, \(i = C\) and \(r = 90^\circ\)):
\[\begin{aligned}
n_1 \sin C &= n_2 \sin 90^\circ \\
(1.33) \sin C &= (1.0)(1) \\
\sin C &= \frac{1}{1.33} \\
C &= \sin^{-1}\left(\frac{1}{1.33}\right) \\
\boldsymbol{C} &= \boldsymbol{48.8^\circ}
\end{aligned}\]
✅ \(C = 48.8^\circ\)
Ex 14.4 Power of Concave Lens
A concave lens has a focal length of \(-25\) cm. Find its power.
Given Data
\(Focal\ length\)\(= f = -25 \, \text{cm} = -0.25 \, \text{m}\)
To Find
\(Power\ of\ lens = P = ?\)
Solution
By using formula of power:
\[\begin{aligned}
P &= \frac{1}{f \, (\text{in metres})} \\
P &= \frac{1}{-0.25} \\
\boldsymbol{P} &= \boldsymbol{-4 \, \text{D}}
\end{aligned}\]
Negative sign shows that it is a concave lens.
✅ \(P = -4 \, \text{D}\)
Ex 14.5 Magnification and Image Height
An object 8 cm tall is placed in front of a concave mirror. The linear magnification is \(-2\). Find the height of the image.
Given Data
\(Object\ height\)\(= h_o = 8 \, \text{cm}\)
\(Magnification\)\(= M = -2\)
To Find
\(Image\ height = h_i = ?\)
Solution
By using magnification formula:
\[\begin{aligned}
M &= \frac{h_i}{h_o} \\
h_i &= M \times h_o \\
h_i &= -2 \times 8 \\
\boldsymbol{h_i} &= \boldsymbol{-16 \, \text{cm}}
\end{aligned}\]
Negative sign shows that the image is inverted and twice the size of the object.
✅ \(h_i = -16 \, \text{cm}\) (Inverted, 2×)
📐 Key Formulas – Light
Refractive Index: \( n = \frac{c}{v} \)
Snell's Law: \( n_1 \sin i = n_2 \sin r \)
Critical Angle: \( C = \sin^{-1}\left(\frac{1}{n}\right) \)
Mirror Formula: \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)
Lens Formula: \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)
Magnification: \( M = \frac{h_i}{h_o} = \frac{|q|}{p} \)
Power of Lens: \( P = \frac{1}{f \, (\text{in metres})} \)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13Sound
Unit 14✨ Light
Unit 15Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and sign conventions (especially for mirrors and lenses). These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus