MP Board · Class 10 · Science · The Human Eye and Colourful WorldA person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type and power of the corrective lens used to restore proper vision? Show complete step-by-step calculations.

To find the type and power of the corrective lens, we need to analyze the given data for the myopic eye:

1. Identify the given values:

   • The far point of the normal human eye is at infinity ($u = \infty$).
   • The far point of the given myopic person is given as $1.2$ m in front of the eye.
   • Therefore, the corrective lens must form a virtual image of a distant object (placed at infinity) at the person's far point ($1.2$ m).
   • Object distance, $u = -\infty$
   • Image distance, $v = -1.2\text{ m} = -120\text{ cm}$ (using sign convention, distances measured in front of the lens are negative).

2. Calculate the focal length ($f$) using the Lens Formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ Substitute the values of $v$ and $u$ into the formula: $$\frac{1}{f} = \frac{1}{-1.2} - \frac{1}{-\infty}$$ Since $\frac{1}{-\infty} = 0$: $$\frac{1}{f} = \frac{1}{-1.2}$$ $$f = -1.2\text{ m}$|

3. Determine the type of lens:

   • Since the calculated focal length $f$ is negative, the corrective lens required is a concave lens (diverging lens).

4. Calculate the Power ($P$) of the Lens:

   • The formula for power is: $$P = \frac{1}{f(\text{in meters})}$$
   • Substitute $f = -1.2\text{ m}$: $$P = \frac{1}{-1.2}$$ $$P = -\frac{10}{12} = -\frac{5}{6}\text{ D}$$ $$P \approx -0.83\text{ D}$$

Conclusion:\nThe person should use a concave lens of power $-0.83\text{ D}$ (or focal length $-1.2\text{ m}$) to correct the myopia defect.