JEE Challenger
More from Units and Measurements
PhyMedAdv 2019 • Paper 2

Percentage Error in Lens Focal Length Measurement on Optical Bench

An optical bench is equipped with a 1.5 m1.5\text{ m} long scale with four equal divisions per cm. In an experiment to determine the focal length of a convex lens, the lens is positioned at the 75 cm75\text{ cm} mark while the object pin is placed at the 45 cm45\text{ cm} mark. The image pin on the opposite side of the lens is adjusted to 135 cm135\text{ cm} mark where it overlaps with the image of the object pin. Determine the percentage error in measuring the focal length of the lens.

Official Numerical Answer1.39

Topics & Concepts

Units and MeasurementsMeasuring InstrumentsLenses and Mirrors

Step-by-Step Solution

To find the percentage error in measuring the focal length of the convex lens, we follow these steps:

1. Least Count of the Scale

The scale on the optical bench has 4 equal divisions in each centimeter. Therefore, the least count (LC\text{LC}) of the scale is: LC=1 cm4=0.25 cm\text{LC} = \frac{1\text{ cm}}{4} = 0.25\text{ cm}

2. Calculation of Object Distance (uu) and Image Distance (vv)

Given positions on the scale:

  • Position of Lens, xL=75 cmx_L = 75\text{ cm}
  • Position of Object Pin, xO=45 cmx_O = 45\text{ cm}
  • Position of Image Pin, xI=135 cmx_I = 135\text{ cm}

The magnitude of the object distance (uu) is: u=∣xL−xO∣=75−45=30 cmu = |x_L - x_O| = 75 - 45 = 30\text{ cm}

The magnitude of the image distance (vv) is: v=∣xI−xL∣=135−75=60 cmv = |x_I - x_L| = 135 - 75 = 60\text{ cm}

3. Calculation of Focal Length (ff)

Using the lens formula: 1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

f=uvu+v=30×6030+60=180090=20 cmf = \frac{uv}{u + v} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\text{ cm}

4. Errors in uu and vv

Since u=xL−xOu = x_L - x_O and v=xI−xLv = x_I - x_L, each measurement involves reading two marks on the scale. Thus, the absolute maximum error in measuring uu and vv is twice the least count: Δu=ΔxL+ΔxO=0.25 cm+0.25 cm=0.50 cm\Delta u = \Delta x_L + \Delta x_O = 0.25\text{ cm} + 0.25\text{ cm} = 0.50\text{ cm} Δv=ΔxI+ΔxL=0.25 cm+0.25 cm=0.50 cm\Delta v = \Delta x_I + \Delta x_L = 0.25\text{ cm} + 0.25\text{ cm} = 0.50\text{ cm}

5. Calculation of Percentage Error in Focal Length

Differentiating the lens formula: 1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

−dff2=−dvv2−duu2-\frac{\mathrm{d}f}{f^2} = -\frac{\mathrm{d}v}{v^2} - \frac{\mathrm{d}u}{u^2}

Taking maximum error combination: Δff2=Δuu2+Δvv2\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}

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

Substitute the known values: Δff=20(0.5302+0.5602)\frac{\Delta f}{f} = 20 \left( \frac{0.5}{30^2} + \frac{0.5}{60^2} \right)

Δff=20×0.5(1900+13600)\frac{\Delta f}{f} = 20 \times 0.5 \left( \frac{1}{900} + \frac{1}{3600} \right)

Δff=10(4+13600)=10×53600=172\frac{\Delta f}{f} = 10 \left( \frac{4 + 1}{3600} \right) = 10 \times \frac{5}{3600} = \frac{1}{72}

The percentage error in focal length is: Percentage Error=Δff×100%=172×100%=2518%≈1.39%\text{Percentage Error} = \frac{\Delta f}{f} \times 100\% = \frac{1}{72} \times 100\% = \frac{25}{18}\% \approx 1.39\%

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