Solveeit Logo
Login

Question

Question: An object is placed at a distance $u = (60 \pm 0.6)$ cm from a mirror. The image is formed on the sa...

An object is placed at a distance u=(60±0.6)u = (60 \pm 0.6) cm from a mirror. The image is formed on the same side of the mirror at a distance v=(30±0.3)v = (30 \pm 0.3) cm.

What is the focal length ff of the mirror, along with its uncertainty?

A

(20±0.4)(20 \pm 0.4) cm

B

(20±0.2)(20 \pm 0.2) cm

C

(20±0.1)(20 \pm 0.1) cm

D

(20±0.6)(20 \pm 0.6) cm

Answer

(20±0.2)(20 \pm 0.2) cm

Explanation

Solution

To find the focal length ff and its uncertainty Δf\Delta f, we use the mirror formula and error propagation.

  1. Mirror Formula:

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

    Given u=−60u = -60 cm and v=−30v = -30 cm (both negative because they are on the same side, using sign convention):

    1f=1−60+1−30=−160−260=−360=−120\frac{1}{f} = \frac{1}{-60} + \frac{1}{-30} = -\frac{1}{60} - \frac{2}{60} = -\frac{3}{60} = -\frac{1}{20}

    f=−20f = -20 cm. The magnitude is ∣f∣=20|f| = 20 cm.

  2. Uncertainty Calculation:

    We use the maximum error method:

    Δf=f2(Δuu2+Δvv2)\Delta f = f^2 \left(\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right)

    Given Δu=0.6\Delta u = 0.6 cm and Δv=0.3\Delta v = 0.3 cm:

    Δf=(20)2(0.6(60)2+0.3(30)2)=400(0.63600+0.3900)\Delta f = (20)^2 \left(\frac{0.6}{(60)^2} + \frac{0.3}{(30)^2}\right) = 400 \left(\frac{0.6}{3600} + \frac{0.3}{900}\right)

    Δf=400(0.63600+1.23600)=400(1.83600)=400×1.83600=7203600=15=0.2\Delta f = 400 \left(\frac{0.6}{3600} + \frac{1.2}{3600}\right) = 400 \left(\frac{1.8}{3600}\right) = \frac{400 \times 1.8}{3600} = \frac{720}{3600} = \frac{1}{5} = 0.2 cm

Therefore, the focal length ff with its uncertainty Δf\Delta f is (20±0.2)(20 \pm 0.2) cm.