A lamp and a screen are set up 100 cm apart and a convex lens is placed between them. The two positions of the lens forming real images on the screen are 40 cm apart. What is the focal length of the lens ?a)21 cmb)15 cmc)18 cmd)12 cmCorrect answer is option 'A'. Can you explain this answer?

Question

Answer

Solution:

Given:

Distance between the lamp and the screen, d = 100 cm
Distance between the two positions of the lens forming real images, d' = 40 cm

To find: Focal length of the lens, f

Using the lens formula,

1/f = 1/v - 1/u

where,
v = distance of image from the lens
u = distance of object from the lens

Let the first position of the lens be at a distance x from the lamp and the second position be at a distance (d-x) from the lamp.

Using the lens formula for both positions,

1/f = 1/v1 - 1/u1 ...(1)
1/f = 1/v2 - 1/u2 ...(2)

where,
u1 = x
v1 = d - x - f
u2 = d - x
v2 = x + f

Substituting the values of u1, v1, u2, and v2 in equations (1) and (2), we get

1/f = (d - x - f - 100)/(d - x) - x/(d - x) ...(3)
1/f = (x + f - 100)/x - (d - x)/x ...(4)

Multiplying equations (3) and (4), we get

1/f^2 = [(d - x - f - 100)/(d - x) - x/(d - x)][(x + f - 100)/x - (d - x)/x]

Simplifying the above equation, we get

f^2 = (d - x)(x) ...(5)

Using equation (5), we can find the value of x as

x = d/2 ± (d'^2 - 4d^2)/8d

Substituting the values of d, d', we get

x = 30 ± 5/2

Taking x = 30 + 5/2, we get

x = 32.5 cm

Substituting the value of x in equation (5), we get

f^2 = (100 - 32.5)(32.5)

f^2 = 2106.25

f = 21 cm

Therefore, the focal length of the lens is 21 cm.

Answer: (a) 21 cm

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