Question

A person can see objects clearly when they lie between 40 cm and 400 cm from his eye. In order to increase the maximum distance of distant vision to infinity, the type of lens and power of correction lens required respectively are:

• A. Convex, \( 0.25 \) Diopter
• B. Concave, \( -0.25 \) Diopter
• C. Concave, \( -0.5 \) Diopter
• D. Convex, \( 0.5 \) Diopter

Hint:

Myopia (short-sightedness) is corrected using a concave lens. The power of the lens is given by \( P = \frac{1}{f} \), where \( f \) is in meters.

Answer 1

The Correct Option is B

Step 1: Identifying the Eye Defect - The person can see near objects clearly but distant objects appear blurred. - This is a case of myopia (short-sightedness). - A concave lens is required to correct this.
Step 2: Calculating the Lens Power The lens formula is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}. \] For distant vision correction: - Far point = \( 400 \) cm. - To correct myopia, the image should be formed at \( 400 \) cm when the object is at infinity (\( u = \infty \)). \[ \frac{1}{f} = \frac{1}{400} - \frac{1}{\infty} = \frac{1}{400}. \] \[ f = 400 \text{ cm} = 4 \text{ m}. \] \[ P = \frac{1}{f} = \frac{1}{-4} = -0.25 \text{ D}. \] Thus, the correct answer is: \[ \boxed{\text{Concave, } -0.25 \text{ D}}. \]