For a normal eye, the cornea of eye provides a converging power of $40 ext{ D}$ and the least converging powe

Question

For a normal eye, the cornea of eye provides a converging power of $40	ext{ D}$ and the least converging power of the eye lens behind the cornea is $20	ext{ D}$. Using this information, the distance between the retina and the cornea-eye lens can be estimated to be:

Accepted Answer

1.  **Total Converging Power ($P$):** The effective power of the eye is the sum of the power of the cornea and the power of the eye lens (assuming they act as a combined lens system close to each other).
    *   $P = P_{cornea} + P_{lens}$
    *   $P = 40	ext{ D} + 20	ext{ D} = 60	ext{ D}$.
2.  **Focal Length ($f$):** The relationship between power (in Diopters) and focal length (in meters) is $P = \frac{1}{f}$.
    *   $f = \frac{1}{P} = \frac{1}{60}	ext{ m}$.
3.  **Unit Conversion:** Convert the focal length from meters to centimeters.
    *   $f = \frac{100}{60}	ext{ cm} = \frac{10}{6}	ext{ cm} \approx 1.67	ext{ cm}$.
4.  **Distance to Retina:** For a normal eye viewing a distant object (least converging power), the image forms on the retina. Therefore, the distance between the optical center of the eye system and the retina is equal to the focal length, which is approximately $1.67	ext{ cm}$.

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