Question 15 Marks
a. What is a lens? List two main categories of lenses. In which category is a double concave lens placed?
b. A convex lens of focal length 15 cm forms a real image at a distance of 20 cm from its optical centre. Find the position of the object. Is the image formed by the lens magnified or diminished?
b. A convex lens of focal length 15 cm forms a real image at a distance of 20 cm from its optical centre. Find the position of the object. Is the image formed by the lens magnified or diminished?
Answer
View full question & answer→a. A transparent material bounded by two surfaces of which one or both surfaces are spherical / curved.
a. Converging lens
b. Diverging lens
Double concave lens is a diverging lens.
b. To find the position of the object, we can use the lens formula:
$\mathrm{f}=15 \mathrm{~cm}, \mathrm{v}=20 \mathrm{~cm}, \mathrm{u}=$ ?
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$
$=\frac{1}{20 \mathrm{~cm}}-\frac{1}{15 \mathrm{~cm}}$
$=\frac{4-3}{60}$
$\mathrm{u}=-60 \mathrm{~cm}$.
image is diminished.
a. Converging lens
b. Diverging lens
Double concave lens is a diverging lens.
b. To find the position of the object, we can use the lens formula:
$\mathrm{f}=15 \mathrm{~cm}, \mathrm{v}=20 \mathrm{~cm}, \mathrm{u}=$ ?
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$
$=\frac{1}{20 \mathrm{~cm}}-\frac{1}{15 \mathrm{~cm}}$
$=\frac{4-3}{60}$
$\mathrm{u}=-60 \mathrm{~cm}$.
image is diminished.

