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Physics: Space and Astrophysics - Curved Lenses, Image Formation, and the Human Eye

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A convex lens is a converging lens that is thicker in the center than at the edges. It converges parallel rays of light to a single point called the principal focus (FF).

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A concave lens is a diverging lens that is thinner in the center. It spreads parallel rays of light outward so they appear to originate from a virtual principal focus.

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The focal length (ff) is the distance from the optical center of the lens to the principal focus. It is related to the radius of curvature (RR) by the approximate relation f=R2f = \frac{R}{2}.

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Images can be Real (can be projected on a screen, formed by actual intersection of rays) or Virtual (cannot be projected, formed by rays appearing to meet).

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The Human Eye acts like a camera; it contains a convex lens that forms a real, inverted image on the retina. The iris controls the amount of light entering through the pupil.

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Accommodation is the ability of the ciliary muscles to change the curvature (and thus the focal length) of the eye lens to focus on objects at various distances.

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Myopia (Short-sightedness) is a defect where distant objects appear blurry because the image forms in front of the retina. It is corrected using a concave lens.

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Hypermetropia (Long-sightedness) is a defect where nearby objects appear blurry because the image forms behind the retina. It is corrected using a convex lens.

📐Formulae

1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

m=hiho=vum = \frac{h_{i}}{h_{o}} = \frac{v}{u}

P=1f (in meters)P = \frac{1}{f \text{ (in meters)}}

f=R2f = \frac{R}{2}

💡Examples

Problem 1:

An object is placed at a distance of 20 cm20\text{ cm} from a convex lens of focal length 10 cm10\text{ cm}. Find the position of the image.

Solution:

Given: u=−20 cmu = -20\text{ cm} (using sign convention), f=+10 cmf = +10\text{ cm}. Using the lens formula: 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} 1v−1−20=110\frac{1}{v} - \frac{1}{-20} = \frac{1}{10} 1v+120=110\frac{1}{v} + \frac{1}{20} = \frac{1}{10} 1v=110−120\frac{1}{v} = \frac{1}{10} - \frac{1}{20} 1v=2−120\frac{1}{v} = \frac{2 - 1}{20} 1v=120\frac{1}{v} = \frac{1}{20} v=20 cmv = 20\text{ cm}

Explanation:

Since vv is positive, the image is formed on the other side of the lens at a distance of 20 cm20\text{ cm}. This is a real and inverted image.

Problem 2:

A person has a lens with power P=−2.0 DP = -2.0\text{ D}. What is the focal length and what type of vision defect does this correct?

Solution:

Given P=−2.0 DP = -2.0\text{ D}. Using the formula P=1fP = \frac{1}{f}: −2.0=1f-2.0 = \frac{1}{f} f=−12.0=−0.5 mf = -\frac{1}{2.0} = -0.5\text{ m} Converting to centimeters: f=−50 cmf = -50\text{ cm}

Explanation:

The focal length is negative, which indicates a concave lens. Therefore, the person is suffering from Myopia (short-sightedness).