JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Geometrical Optics

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In an experiment for estimating the value of focal length of converging mirror, image of an object placed at 40 cm from the pole of the mirror is formed at distance 120 cm from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in 1 cm. The value of error in measurement of focal length of the mirror is

1/ K

cm. The value of

K

____

Answer: 32

Solution

With the mirror formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

and

u = - 40 cm

v = - 120 cm

, the focal length is

f = - 30 cm

. Differentiating gives

\frac{df}{f ^{2}} = \frac{d v}{v ^{2}} + \frac{d u}{u ^{2}}

, and each distance carries the least count

\frac{1}{20} cm

of the

20

-division scale. Substituting,

df = (30)^{2} \times \frac{1}{20} (\frac{1}{12 0 ^{2}} + \frac{1}{4 0 ^{2}}) = 900 \times \frac{1}{20} \times \frac{1}{1440} = \frac{1}{32} cm

. The error in the focal length is

\frac{1}{32} cm

, so

K = 32

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Geometrical Optics

Solution

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