Question

In an experiment to measure the focal length ($f$) of a concave mirror, the object distance ($x$) and image distance ($y$) measured as (24.0 $\pm$ 0.1) cm and (12.0 $\pm$ 0.1) cm respectively, then ($\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{1}{f}$)

• A. Absolute error in focal length is 0.2 cm
• B. Percentage error in focal length is 1.7%
• C. Percentage error in focal length is 0.7%
• D. Percentage error in measurement of object distance is 0.4%

Answer 1

Answer: (C), (D)

C, D

Answer 2

1. Calculate the focal length ($f$):

   The given mirror formula is $\frac{1}{f} = \frac{1}{x} + \frac{1}{y}$.

   Substitute the given values for $x$ and $y$:

   $x = 24.0$ cm $y = 12.0$ cm

   $\frac{1}{f} = \frac{1}{24.0} + \frac{1}{12.0}$

   To add the fractions, find a common denominator (24):

   $\frac{1}{f} = \frac{1}{24} + \frac{2}{24}$ $\frac{1}{f} = \frac{3}{24}$ $\frac{1}{f} = \frac{1}{8}$ $f = 8$ cm

2. Calculate the absolute error in focal length ($\Delta f$):

   To find the absolute error, we use the error propagation formula for the given relation $\frac{1}{f} = \frac{1}{x} + \frac{1}{y}$.

   Differentiating this equation with respect to $f$, $x$, and $y$:

   $-\frac{1}{f^2} df = -\frac{1}{x^2} dx - \frac{1}{y^2} dy$

   For maximum possible absolute error, we consider the magnitudes and add them:

   $\frac{\Delta f}{f^2} = \frac{\Delta x}{x^2} + \frac{\Delta y}{y^2}$

   So, $\Delta f = f^2 \left( \frac{\Delta x}{x^2} + \frac{\Delta y}{y^2} \right)$

   Given:

   $x = 24.0$ cm, $\Delta x = 0.1$ cm $y = 12.0$ cm, $\Delta y = 0.1$ cm $f = 8$ cm

   Substitute these values into the formula for $\Delta f$:

   $\Delta f = (8)^2 \left( \frac{0.1}{(24)^2} + \frac{0.1}{(12)^2} \right)$ $\Delta f = 64 \left( \frac{0.1}{576} + \frac{0.1}{144} \right)$

   To add the fractions, find a common denominator (576):

   $\Delta f = 64 \left( \frac{0.1}{576} + \frac{0.1 \times 4}{144 \times 4} \right)$ $\Delta f = 64 \left( \frac{0.1}{576} + \frac{0.4}{576} \right)$ $\Delta f = 64 \left( \frac{0.1 + 0.4}{576} \right)$ $\Delta f = 64 \left( \frac{0.5}{576} \right)$ $\Delta f = \frac{64 \times 0.5}{576}$ $\Delta f = \frac{32}{576}$ $\Delta f = \frac{1}{18}$ cm

   As a decimal, $\Delta f \approx 0.0555...$ cm.

   Rounding to one significant figure for error, $\Delta f \approx 0.06$ cm.

   Rounding to one decimal place, $\Delta f \approx 0.1$ cm.

   Option A states the absolute error is 0.2 cm, which is incorrect.

3. Calculate the percentage error in focal length:

   Percentage error in $f = \left( \frac{\Delta f}{f} \right) \times 100\%$ Percentage error in $f = \left( \frac{1/18}{8} \right) \times 100\%$ Percentage error in $f = \left( \frac{1}{18 \times 8} \right) \times 100\%$ Percentage error in $f = \left( \frac{1}{144} \right) \times 100\%$ Percentage error in $f = \frac{100}{144} \%$ Percentage error in $f = \frac{25}{36} \%$ Percentage error in $f \approx 0.6944... \%$

   Rounding to one decimal place, the percentage error in focal length is $0.7 \%$.

   Option B states 1.7%, which is incorrect.

   Option C states 0.7%, which is correct.

4. Calculate the percentage error in object distance:

   Percentage error in $x = \left( \frac{\Delta x}{x} \right) \times 100\%$ Percentage error in $x = \left( \frac{0.1}{24.0} \right) \times 100\%$ Percentage error in $x = \frac{10}{24} \%$ Percentage error in $x = \frac{5}{12} \%$ Percentage error in $x \approx 0.4166... \%$

   Rounding to one decimal place, the percentage error in object distance is $0.4 \%$.

   Option D states 0.4%, which is correct.

Therefore, options C and D are correct.