Question

Suppose you find a magnifying glass in a junk drawer at home. You place it directly in between a light bulb and the wall, which are separated by 95 cm. When the lens is 15 cm from the wall it creates a focused image of the bulb. (a) Determine the focal length of the magnifying glass lens. (b) Without moving the light bulb (or the wall) at what other location would the lens project a focused image of the bulb on the wall? (c) Find the ratio of the image sizes of the two cases. (d) If the bulb is moved too close to the wall there will be no way possible to project a focused image – determine this distance and explain.

Answer 1

Ans. (a):

The bulb is the object and its image will be formed on the wall

we have, u + v = 95 cm

As distance between lens and wall is 15 cm and the image is obtained on the wall,

we have v = 15 cm and u = 80 cm

Lens equation,  

substitution,

After calculations, we get f = 12.631 cm

Ans. (b):

As the positions of bulb (object) and wall(screen) are fixed, we have u + v = 95 cm

so that

Lens equation,  

substitution,

After simplification, we get the quadratic equation,

The roots of the equation are u = 80 cm or u = 15 cm

When u = 15 cm, v = 80 cm and when u = 80 cm, v = 15 cm

So, for two positions of the lens ( 15 cm from the wall and 80 cm from the wall) there'll be clear image of the bulb formed on the wall.

Ans. (c):

Let M₁ and M₂ are magnifications obtained in first and second case respectively.

Ans. (d):

When the object is at the focus, the image is formed at infinity.

When the object is between focus and pole of the lens, we get a virtual image which cannot be obtained on a screen (or a wall).

For given lens, if the bulb is at distance of 12.63 cm or less, from the lens, there'll be no image formed on the wall for any distance between the lens and wall.