Question

In displacement method, there are two position of lens for which we get clear image. First position of the lens is at 40 cm from object and second is at 80 cm, the focal length of lens is
(1) 40 cm
(2) 40/3 cm
(3) 80 cm
(4) 80/3 cm

Answer 1

For first lens:

Object distance from the lens i.e. u is -40 cm.

Let us assume that the image formation of the first lens takes place at the screen which is 'x' cm away.

Using lens formula,

1/f = 1/v - 1/u

Substitute the values,

1/f = 1/x - 1/(-40)

1/f = 1/x + 1/40

1/f = (40 + x)/40x

f = 40x/(40 + x)

40f + fx = 40x

40f = 40x - fx

40f = x(40 - f)

40f/(40 - f) = x ...............(1st equation)

For second lens:

Object distance i.e. u from the screen is -80 cm and again the image formation i.e. v is 'x - 40' cm.

Since the first lens image formation was x cm always from the lens and for the second one object is 80 cm away. So, object will be place at image distance from lens - object distance from the first lens. i.e. (x - 40)

Using lens formula,

1/f = 1/v - 1/u

Substitute the values,

1/f = 1/(x-40) - 1/(-80)

1/f = 1/(x-40) + 1/80 ...............(2nd equation)

Substitute value of (1st equation) in (2nd equation)

→ 1/f = 1/[40f/(40-f) - 40] + 1/80

→ 1/f = 1/[(40f - 1600 + 40f)/(40 - f)] + 1/80

→ 1/f = 1/[(80f -1600)/(40 - f)] + 1/80

→ 1/f = (40 - f)/(80f -1600) + 1/80

→ 1/f = 1/80 × (40 - f)/(f - 20) + 1/1

→ 1/f = 1/80 × (40 - f + f - 20)/(f - 20)

→ 1/f = 1/80 × (20)/(f - 20)

→ 1/f = 20/80 × 1/(f - 20)

→ 1/f = 1/4 × 1/(f - 20)

→ 4/f = 1/(f - 20)

Cross-multiply them,

→ 4f - 80 = f

→ 4f - f = 80

→ 3f = 80

→ f = 80/3

Therefore, the focal length of the lens is 80/3 cm.

Option (4) 80/3 cm

Answer 2

SOLUTION :-

=> For first lens,

=> Given that, object distance from lens is -40cm .

since, u = -40cm.

=> Let assume that, the image formation of the first lens takes place at the screen which is 'r'cm.

=>Now, we have value of v and u. this value put in lens formula, and we get,

=> 1/f = 1/v- 1/u

=> 1/f = 1/r - 1/-40

[After solving, we get ]

=> r = 40f/(40-f)

=> For second lens,

=>Given that, object distance from lens is -40cm .

since, u = -80cm.

=> Let assume that, image formation from second lens i.e, 'r-40cm'.

=>Now, we have value of v and u. this value put in lens formula, and we get,

=> 1/f = 1/v - 1/u

=>1/f = 1/(r-40) - 1/-80

=> .°. 1/f = 1/(r-40)+ 1/80--------(1)

[Put r value in equation. (1)]

=>1/f = 1/[(40f/(40-f)]+1/80

[After solving, we get]

=>4/f = 1/(f - 20)

=> 4(f-20)=1(f) ---------(Apply cross multiplication)

=> 4f - 80 = f

=> 4f - f = 80

=> 3f = 80

=> .°. f = 80/3.

So, the focal length of lens is 80/3cm.

Therefore, the correct option is 4.

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