Solveeit Logo

Question

Question: An electric lamp and a screen are placed on the table, in a line at a distance of \(1m\). In what po...

An electric lamp and a screen are placed on the table, in a line at a distance of 1m1m. In what position of a convex lens of the focal length of f=21cmf = 21cm will the image of the lamp be sharp?

Explanation

Solution

A convex lens converges the lights coming from the source. Here the source is the light bulb which is at a fixed distance from the screen. To get a sharp image on the screen, the image should be formed just on the screen. To get it on the screen, the image should be real.

Formula Used:
If a lens has a focal length ff and if the object is placed at a distance uu and the image is formed at a distance vv then the relation is
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

Complete step by step answer:
Given, the focal length of the lens is 21cm21cm.
The screen and the bulb have a distance of 1m1m or 100cm100cm.
To get: The position to get a sharp image of the bulb.

Step 1:

Since the image is real, the screen and the lamp should be on opposite sides of the convex lens.
Let the distance from the object to the lens is xx cm.
So, the distance from the lens to the screen is (100x)cm\left( {100 - x} \right)cm.

Step 2:
You now have, u=xcmu = - xcm, v=(100x)cmv = \left( {100 - x} \right)cm and f=21cmf = 21cm
Compute the value of xx from the relation in eq (1)

121=1100x+1x 121=x+100x100xx2 100xx2=21x+210021x x2100x2100=0 (x30)×(x70)=0  \dfrac{1}{{21}} = \dfrac{1}{{100 - x}} + \dfrac{1}{x} \\\ \Rightarrow \dfrac{1}{{21}} = \dfrac{{x + 100 - x}}{{100x - {x^2}}} \\\ \Rightarrow 100x - {x^2} = 21x + 2100 - 21x \\\ \Rightarrow {x^2} - 100x - 2100 = 0 \\\ \Rightarrow \left( {x - 30} \right) \times \left( {x - 70} \right) = 0 \\\

Hence, the solutions are x=30cmx = 30cm or x=70cmx = 70cm .

The lens should be placed in the middle of the screen at a distance of 30cm30cm or 70cm70cm from the bulb.

Note:
The distance of the object from the lens in the eq (1) should be negative as the direction of light is different. The image should be real so that it can be projected on the screen, hence the screen and the object should be on the opposite side of the lens. You must be careful in these two points. Then from the relation, you shall get a quadratic equation in this type of problem. Hence, you will have two solutions that you can check for validity with other conditions given.