Question

Solve the following examples.
i. Doctor has prescribed a lens having power +1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
ii. 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
iii. Three lenses having power 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?
iv. An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?

Answer 1

i. Given:
Power of lens, P = +1.5 D
Now, focal length of lens, $f=\frac{1}{P}=+\frac{1}{1.5}=+0.67\mathrm{m}$.
Since, the focal length is positive, the lens prescribed for correction is convex lens. Thus, the defect of vision is farsightedness or hypermetropia.

ii. Given:
Height of object, h_o = 5 cm
Object distance, u = $-$25 cm
Since the lens is converging, thus it is a convex lens.
Focal length of the lens, f = 10 cm
Using lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=\frac{1}{10}+\frac{1}{-25}=\frac{3}{50} \\ \Rightarrow v=\frac{50}{3}=16.7\mathrm{cm} \end{gathered}$$
Thus, the image is formed $16.7\mathrm{cm}$ right of the lens.
Now, we know
$$\begin{gathered} \frac{v}{u}=\frac{h_{\mathrm{i}}}{h_{\mathrm{o}}} \\ \Rightarrow h_{\mathrm{i}}=\frac{50}{3\times -25}\times 5=\frac{10}{3}=-3.3\mathrm{cm} \end{gathered}$$
Thus, the size of the image is 3.3 cm. Negative sign shows that the image formed is real and inverted. Hence, the image formed is real and inverted and diminished.

iii. Given:
P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D
Let the total power of the lens combination be P. Thus,
$P=P_1+P_2+P_3=2+2.5+1.7=6.6\mathrm{D}$

iv. Given:
Object distance, u = $-$60 cm
Image distance, v = $-$20 cm
Using lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{f}=\frac{{\displaystyle 1}}{{\displaystyle -20}}-\frac{{\displaystyle 1}}{{\displaystyle -60}}=-\frac{1}{30} \\ \Rightarrow f=-30\mathrm{cm} \end{gathered}$$
Since, the focal length is negative, the lens is a diverging lens or a concave lens.