Class Assignment 2 – Lenses
Conceptual Questions
1. What is a lens? How many types of lenses are there?A lens is a transparent optical device bounded by two curved surfaces that refracts light. There are two main types of lenses:
- Convex lens
- Concave lens
(a) Convex lens: It is thicker at the centre and thinner at the edges. It converges parallel rays of light to a point called the focus.
(b) Concave lens: It is thinner at the centre and thicker at the edges. It diverges parallel rays of light.
3. Distinguish between a convex lens and a concave lens.
| Convex Lens | Concave Lens |
|---|---|
| Thicker at the centre | Thinner at the centre |
| Converges light rays | Diverges light rays |
| Can form real and virtual images | Always forms virtual image |
4. Why is a convex lens called a converging lens?
A convex lens bends parallel rays of light towards a single point called the principal focus. Therefore it is called a converging lens.
5. Why is a concave lens called a diverging lens?
A concave lens spreads or diverges parallel rays of light after refraction. Hence it is called a diverging lens.
6. A thin lens has a focal length \(f=-12\) cm. Is it convex or concave?
Negative focal length indicates a concave lens.
7. If the image formed by a lens is always diminished and erect, what is the nature of the lens?
Such images are produced only by a concave lens.
8. For what position of an object is a virtual image formed by a convex lens?
When the object is placed between the optical centre and the focus (distance less than f).
9. Where should an object be placed so that a real and inverted image of the same size is obtained using a convex lens?
The object should be placed at 2f.
10. If the image formed by a convex lens is the same size as the object, where is the image formed?
The image is formed at 2f on the opposite side of the lens.
11. A 1 cm high object is placed at a distance of 2f from a convex lens. Find the height of the image.
At object distance = 2f, image is same size but inverted. \[ h_i = -1 \text{ cm} \] Negative sign shows the image is inverted.
Numerical Problems
1. Object height = 5 cm, object distance = 10 cm, focal length = 6 cm \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] \[ \frac{1}{6}=\frac{1}{v}-\frac{1}{(-10)} \] \[ \frac{1}{6}=\frac{1}{v}+\frac{1}{10} \] \[ \frac{1}{v}=\frac{1}{6}-\frac{1}{10} \] \[ v=15\,cm \] Magnification: \[ m=\frac{v}{u}=\frac{15}{-10}=-1.5 \] \[ h_i=m\times h_o=-1.5\times5=-7.5\,cm \] Answer: Image at 15 cm, real, inverted, magnified, height = 7.5 cm.2. Focal length = 15 cm, object distance = 30 cm \[ \frac{1}{15}=\frac{1}{v}-\frac{1}{(-30)} \] \[ \frac{1}{15}=\frac{1}{v}+\frac{1}{30} \] \[ v=30\,cm \] Magnification: \[ m=\frac{30}{-30}=-1 \] Answer: Image at 30 cm, real, inverted, same size.
3. Camera lens focal length = 10 cm (a) Object at infinity \[ v=f=10\,cm \] (b) Object at 20 cm \[ \frac{1}{10}=\frac{1}{v}-\frac{1}{(-20)} \] \[ \frac{1}{10}=\frac{1}{v}+\frac{1}{20} \] \[ v=20\,cm \] Answer: (a) 10 cm (b) 20 cm
4. Object height = 4 cm, object distance = 15 cm, focal length = 5 cm \[ \frac{1}{5}=\frac{1}{v}-\frac{1}{(-15)} \] \[ \frac{1}{5}=\frac{1}{v}+\frac{1}{15} \] \[ v=7.5\,cm \] Magnification: \[ m=\frac{7.5}{-15}=-0.5 \] \[ h_i=-0.5\times4=-2\,cm \] Answer: Image at 7.5 cm, height = 2 cm, real and inverted.