Class 10 Physics – Lenses Numerical Practice (Set 1)
Important Sign Convention for Lenses
- Light is assumed to travel from left to right.
- Distances measured opposite to incident light → negative.
- Distances measured along incident light → positive.
- 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.
Question 1
An object is placed 30 cm in front of a convex lens of focal length 20 cm.
Find the image distance and nature of the image.
Hint: Convex lens → \(f\) positive.
Object in front → \(u\) negative.
Solution
\[
f = +20\,cm
\]
\[
u = -30\,cm
\]
Lens formula
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
\[
\frac{1}{20}=\frac{1}{v}-\frac{1}{(-30)}
\]
\[
\frac{1}{20}=\frac{1}{v}+\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{1}{20}-\frac{1}{30}
\]
LCM = 60
\[
\frac{1}{v}=\frac{3-2}{60}
\]
\[
v=60\,cm
\]
Image is real and inverted.
Question 2
An object is placed 10 cm in front of a convex lens of focal length 15 cm.
Find the image distance.
Hint: Object is inside focus → image becomes virtual.
Solution
\[
f=+15
\]
\[
u=-10
\]
\[
\frac{1}{15}=\frac{1}{v}-\frac{1}{(-10)}
\]
\[
\frac{1}{15}=\frac{1}{v}+\frac{1}{10}
\]
\[
\frac{1}{v}=\frac{1}{15}-\frac{1}{10}
\]
\[
\frac{1}{v}=\frac{2-3}{30}
\]
\[
v=-30\,cm
\]
Negative \(v\) means image is virtual.
Question 3
A concave lens has focal length 15 cm.
An object is placed 20 cm in front of the lens.
Find image distance.
Hint: Concave lens → focal length negative.
Solution
\[
f=-15
\]
\[
u=-20
\]
\[
\frac{1}{-15}=\frac{1}{v}-\frac{1}{(-20)}
\]
\[
\frac{1}{-15}=\frac{1}{v}+\frac{1}{20}
\]
\[
\frac{1}{v}=-\frac{1}{15}-\frac{1}{20}
\]
\[
\frac{1}{v}=-\frac{7}{60}
\]
\[
v=-8.6\,cm
\]
Image is virtual and erect.
Question 4
Find the magnification when image distance = 40 cm and object distance = 20 cm.
Solution
\[
m=\frac{v}{u}
\]
\[
m=\frac{40}{-20}
\]
\[
m=-2
\]
Image is inverted.
Question 5
A convex lens produces an image at 30 cm when the object is placed 15 cm away.
Find focal length.
Solution
\[
u=-15
\]
\[
v=+30
\]
\[
\frac{1}{f}=\frac{1}{30}-\frac{1}{-15}
\]
\[
\frac{1}{f}=\frac{1}{30}+\frac{1}{15}
\]
\[
\frac{1}{f}=\frac{3}{30}
\]
\[
f=10\,cm
\]
Question 6
An object is placed at 25 cm from a convex lens of focal length 10 cm.
Find image distance.
Solution
\[
f=10
\]
\[
u=-25
\]
\[
\frac{1}{10}=\frac{1}{v}+\frac{1}{25}
\]
\[
\frac{1}{v}=\frac{1}{10}-\frac{1}{25}
\]
\[
\frac{1}{v}=\frac{3}{50}
\]
\[
v=16.7\,cm
\]
Question 7
A concave lens of focal length 20 cm forms an image at 10 cm.
Find object distance.
Solution
\[
f=-20
\]
\[
v=-10
\]
\[
\frac{1}{-20}=\frac{1}{-10}-\frac{1}{u}
\]
\[
\frac{1}{u}=-\frac{1}{20}
\]
\[
u=-20\,cm
\]
Question 8
Object height = 2 cm
Magnification = 3
Find image height.
Solution
\[
m=\frac{h_i}{h_o}
\]
\[
3=\frac{h_i}{2}
\]
\[
h_i=6\,cm
\]
Question 9
Object is placed at 2f from convex lens.
Describe image.
Answer
Image forms at 2f
Real
Inverted
Same size
Question 10
Why does a concave lens always produce virtual image?
Answer
Because the refracted rays diverge and appear to meet behind the lens.
Hence the image is always virtual and erect.
