CBSE · Class 10 · Science · The Human Eye and Colourful WorldA person with a myopic eye cannot see objects beyond 1.2 m distinctly. (a) What is the defect the person is suffering from? (b) What is the cause of this defect? (c) Calculate the focal length and power of the corrective lens required to restore proper vision.

Solution:

(a) Identification of Defect:\nThe person is suffering from Myopia (near-sightedness), a vision defect where an individual can see nearby objects clearly, but distant objects appear blurred because the far point of the eye has reduced from infinity to a closer distance (1.2 m in this case).

(b) Causes of Myopia:\nMyopia is typically caused by:

1. Excessive curvature of the eye lens.
2. Elongation of the eyeball.

(c) Calculation of Focal Length and Power:\nGiven data:

• The far point of the defective eye ($D$) = $-1.2 \text{ m} = -120 \text{ cm}$ (since the object distance for the distant object is at infinity, $u = -\infty$, but the lens must form the virtual image of an object at infinity at the person's far point).
• Therefore, Image distance ($v$) = $-1.2 \text{ m}$
• Object distance ($u$) = $-\infty$ \nUsing the Lens Formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ \nSubstituting the values: $$\frac{1}{f} = \frac{1}{-1.2} - \frac{1}{-\infty}$| $$\frac{1}{f} = \frac{1}{-1.2} + 0$$ $$\frac{1}{f} = -\frac{1}{1.2}$$ $$f = -1.2 \text{ m}$$ \nThus, the focal length of the corrective concave lens is $-1.2 \text{ m}$ (or $-120 \text{ cm}$). \nNow, calculating the Power ($P$) of the lens: $$P = \frac{1}{f \text{ (in meters)}}$$ $$P = \frac{1}{-1.2}$$ $$P = -\frac{10}{12} = -0.83 \text{ Dioptres (D)}$$

Answer:\nThe focal length of the corrective lens is $-1.2 \text{ m}$ and its power is $-0.83 \text{ D}$ (a concave lens).