focal length & power of lens

Question

A farsighted eye cannot form a sharp image of an object which is closer than 40 cm. Find the focal length & power of a lens which will permit the eye to form a sharp image of an object 25 cm distant, but not an object closer than this.

Answer 1

Identify the Given Values

1. Object Distance (do​): 25 cm (The position where we want to place the object).
2. Image Distance (di​): −40 cm.
3. Note: The image distance is negative because the lens must form a virtual image on the same side as the object. This virtual image serves as the “object” that the eye can focus on at its natural limit (40 cm).

2. Calculate the Focal Length (f)

We use the thin lens equation:

f1​=do​1​+di​1​

Substitute the values:

f1​=251​+−401​

Find a common denominator (200):

f1​=2008​−2005​

f1​=2003​

Solve for f:

f=3200​ cm

f≈66.7 cm

The focal length is approximately 66.7 cm.

3. Calculate the Power (P)

The power of a lens is calculated using the formula P=f1​, where f must be in meters.

Convert focal length to meters:

f=3200​ cm=32​ m≈0.667 m

Calculate Power:

P=0.667 m1​

P=+1.5 Diopters (D)

Final Answer

To correct this vision, a converging (convex) lens is required with:

1. Focal Length: 66.7 cm
2. Power: +1.5 D