Class 10th Lens Question Set-2

Sign Convention Reminder

• Light is assumed to travel from left to 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.

Lens formula: \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] Magnification: \[ m=\frac{v}{u} \]

1. An object is placed 40 cm in front of a convex lens of focal length 20 cm. Find image distance.

Hint: Convex lens → \(f\) positive. Object in front → \(u\) negative.

Show Solution

\[ f=+20\,cm \] \[ u=-40\,cm \] \[ \frac{1}{20}=\frac{1}{v}-\frac{1}{(-40)} \] \[ \frac{1}{20}=\frac{1}{v}+\frac{1}{40} \] \[ \frac{1}{v}=\frac{1}{20}-\frac{1}{40} \] \[ \frac{1}{v}=\frac{1}{40} \] \[ v=40\,cm \] Image is real and inverted.

2. An object is placed 12 cm in front of a convex lens of focal length 18 cm. Find image distance.

Hint: Object inside focus → image virtual.

Show Solution

\[ f=18 \] \[ u=-12 \] \[ \frac{1}{18}=\frac{1}{v}-\frac{1}{(-12)} \] \[ \frac{1}{18}=\frac{1}{v}+\frac{1}{12} \] \[ \frac{1}{v}=\frac{1}{18}-\frac{1}{12} \] \[ \frac{1}{v}=-\frac{1}{36} \] \[ v=-36\,cm \] Image is virtual and erect.

3. A concave lens has focal length 10 cm. An object is placed 15 cm in front of it. Find image distance.

Hint: Concave lens → focal length negative.

Show Solution

\[ f=-10 \] \[ u=-15 \] \[ \frac{1}{-10}=\frac{1}{v}-\frac{1}{(-15)} \] \[ \frac{1}{-10}=\frac{1}{v}+\frac{1}{15} \] \[ \frac{1}{v}=-\frac{1}{10}-\frac{1}{15} \] \[ \frac{1}{v}=-\frac{5}{30} \] \[ v=-6\,cm \] Image is virtual and diminished.

4. An image is formed at 30 cm by a convex lens of focal length 10 cm. Find object distance.

Hint: Real image → \(v\) positive.

Show Solution

\[ v=+30 \] \[ f=+10 \] \[ \frac{1}{10}=\frac{1}{30}-\frac{1}{u} \] \[ \frac{1}{u}=\frac{1}{30}-\frac{1}{10} \] \[ \frac{1}{u}=-\frac{2}{30} \] \[ u=-15\,cm \]

5. A convex lens produces an image twice the size of the object. Object distance = 20 cm. Find image distance.

Hint: Use magnification formula.

Show Solution

\[ m=2 \] \[ m=\frac{v}{u} \] \[ 2=\frac{v}{-20} \] \[ v=-40\,cm \] Image is virtual.

6. Object height = 3 cm. Magnification = −2. Find image height.

Show Solution

\[ m=\frac{h_i}{h_o} \] \[ -2=\frac{h_i}{3} \] \[ h_i=-6\,cm \] Image is inverted.

7. A convex lens forms a real image at 24 cm when object is at 12 cm. Find focal length.

Show Solution

\[ u=-12 \] \[ v=24 \] \[ \frac{1}{f}=\frac{1}{24}-\frac{1}{-12} \] \[ \frac{1}{f}=\frac{1}{24}+\frac{1}{12} \] \[ \frac{1}{f}=\frac{3}{24} \] \[ f=8\,cm \]

8. A concave lens produces an image at 12 cm when object is at 24 cm. Find focal length.

Show Solution

\[ u=-24 \] \[ v=-12 \] \[ \frac{1}{f}=\frac{1}{-12}-\frac{1}{-24} \] \[ \frac{1}{f}=-\frac{1}{24} \] \[ f=-24\,cm \]

9. A convex lens produces an image at infinity. Where is the object placed?

Show Solution

Object is placed at the **principal focus (F)**.

10. If object is placed at infinity in front of a convex lens, where will image form?

Show Solution

Image forms at the **principal focus of the lens**.