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Chapter 18: Problem 157

The distance between an object and the screen is \(100 \mathrm{~cm} .\) A lens produces an image on the screen when placed at either of the positions \(40 \mathrm{~cm}\) apart. The power of the lens is nearly

Short Answer

Expert verified
The power of the lens is approximately \(2.02 \, \mathrm{D}\) (Diopters).

Step by step solution

01

Determining the positions of the lens

The initial position of the lens is \(d = 100 \, \mathrm{cm}\). The two positions of the lens are \(40 \, \mathrm{cm}\) apart, so one position would be \((d - 20) = 80 \, \mathrm{cm}\) and the other position would be \((d + 20) = 120 \, \mathrm{cm}\).
02

Apply the lens formula

Apply the lens formula \(1/f = 1/v - 1/u\) for both positions and solve for \(f\). Since \(v = d\) (distance from the screen) in both scenarios and the object is at infinity, we have to calculate the value of \(f\) twice: first for \(u = -80 \, \mathrm{cm}\) and then for \(u = -120 \, \mathrm{cm}\).
03

Solving for f for the first position

Plugging \(v = 100 \, \mathrm{cm}\) and \(u = -80 \, \mathrm{cm}\) into our lens formula gives us \(1/f = 1/100 + 1/80\). Using a common denominator of 800 to simplify this equation, we get \(f = 44.44 \, \mathrm{cm}\).
04

Solving for f for the second position

Now plug \(v = 100 \, \mathrm{cm}\) and \(u = -120 \, \mathrm{cm}\) into the lens formula, to get \(1/f = 1/100 + 1/120\). Simplifying this equation gives us \(f = 54.54 \, \mathrm{cm}\).
05

Calculate the average focal length

The actual focal length will be the average of these two calculated focal lengths. So \(f = (44.44 + 54.54)/2 = 49.49 \, \mathrm{cm}\) or \(0.4949 \, \mathrm{m}\).
06

Calculate the power of the lens

Power of the lens is the reciprocal of the focal length (measured in meters), or \(P = 1/f\). Substituting the value calculated in Step 5, we find \(P = 1/0.4949 = 2.02 \, \mathrm{D}\) (Diopters). Therefore, the power of this lens is nearly \(2.02 \, \mathrm{D}\).

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