Class 10 Physics – Lenses Numerical Practice (Set 3 – Advanced Level)
Reminder – New Cartesian Sign Convention
- Light travels from left → right.
- Object distance \(u\) → always negative.
- Convex lens focal length \(f\) → positive.
- Concave lens focal length \(f\) → negative.
- Real image distance \(v\) → positive.
- Virtual image distance \(v\) → negative.
1. An object is placed 15 cm in front of a convex lens of focal length 10 cm. Find image distance and magnification.
Hint: Object between F and 2F.
Hint: Object between F and 2F.
Show Solution
\[ f=10 \] \[ u=-15 \] \[ \frac{1}{10}=\frac{1}{v}-\frac{1}{(-15)} \] \[ \frac{1}{10}=\frac{1}{v}+\frac{1}{15} \] \[ \frac{1}{v}=\frac{1}{10}-\frac{1}{15} \] \[ \frac{1}{v}=\frac{1}{30} \] \[ v=30\,cm \] Magnification \[ m=\frac{30}{-15}=-2 \] Image is real and inverted.
2. A convex lens forms a real image at 40 cm when the object is at 20 cm. Find focal length.
Hint: Use lens formula carefully.
Show Solution
\[ u=-20 \] \[ v=40 \] \[ \frac{1}{f}=\frac{1}{40}-\frac{1}{-20} \] \[ \frac{1}{f}=\frac{1}{40}+\frac{1}{20} \] \[ \frac{1}{f}=\frac{3}{40} \] \[ f=13.3\,cm \]
3. A concave lens has focal length 12 cm. Object distance = 24 cm. Find image distance.
Hint: Concave lens → focal length negative.
Show Solution
\[ f=-12 \] \[ u=-24 \] \[ \frac{1}{-12}=\frac{1}{v}-\frac{1}{(-24)} \] \[ \frac{1}{-12}=\frac{1}{v}+\frac{1}{24} \] \[ \frac{1}{v}=-\frac{1}{12}-\frac{1}{24} \] \[ \frac{1}{v}=-\frac{3}{24} \] \[ v=-8\,cm \]
4. Object height = 4 cm. Image height = −8 cm. Object distance = 20 cm. Find image distance.
Hint: First find magnification.
Show Solution
\[ m=\frac{h_i}{h_o} \] \[ m=\frac{-8}{4} \] \[ m=-2 \] \[ -2=\frac{v}{-20} \] \[ v=40\,cm \]
5. Find the power of a convex lens whose focal length is 25 cm.
Show Solution
Convert focal length to metre. \[ f=0.25\,m \] Power \[ P=\frac{1}{f} \] \[ P=4D \]
6. A concave lens has focal length 50 cm. Find its power.
Show Solution
\[ f=-0.5\,m \] \[ P=\frac{1}{f} \] \[ P=-2D \]
7. A convex lens produces an image at infinity. What is object distance?
Show Solution
Object is at **principal focus**.
8. A convex lens of focal length 20 cm forms an image at 60 cm. Find object distance.
Show Solution
\[ f=20 \] \[ v=60 \] \[ \frac{1}{20}=\frac{1}{60}-\frac{1}{u} \] \[ \frac{1}{u}=\frac{1}{60}-\frac{1}{20} \] \[ \frac{1}{u}=-\frac{2}{60} \] \[ u=-30\,cm \]
9. An object placed at 30 cm produces image at 15 cm. Find magnification.
Show Solution
\[ m=\frac{v}{u} \] \[ m=\frac{15}{-30} \] \[ m=-0.5 \]
10. A convex lens forms an image three times the object size. Object distance = 15 cm. Find image distance.