Question 33

What is the relative decrease in focal length of a lens for an increase in optical power  by $$0.1\,D$$ from  $$2.5\,D? \quad [\text{'D' stands for dioptre}] $$

Solution

Share

Initial power is $$P_1 = 2.5$$ D, which gives $$f_1 = 1/2.5 = 0.4$$ m.

When the power increases to $$P_2 = 2.6$$ D, the focal length becomes $$f_2 = 1/2.6$$ m.

The relative decrease is $$\frac{f_1-f_2}{f_1} = 1 - \frac{f_2}{f_1} = 1 - \frac{P_1}{P_2} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} \approx 0.0385 \approx 0.04$$.

Therefore the correct answer is Option 2: 0.04.

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests