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Light: Mirrors and Lenses - What Is a Lens?

Grade 8CBSE

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

🔑Concepts

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A lens is a piece of transparent material (like glass) bounded by two surfaces, at least one of which is a curved surface.

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Convex Lens (Converging Lens): It is thicker in the middle than at the edges. It converges (brings together) parallel rays of light at a point called the Principal Focus (FF).

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Concave Lens (Diverging Lens): It is thinner in the middle than at the edges. It diverges (spreads out) parallel rays of light, making them appear to come from the Principal Focus (FF).

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Optical Centre (OO): The geometric centre of the lens. Light rays passing through OO do not deviate from their path.

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Principal Axis: An imaginary horizontal line passing through the optical centre and the centers of curvature of the lens surfaces.

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Focal Length (ff): The distance between the optical centre (OO) and the principal focus (FF) of the lens.

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A convex lens can form both real (inverted) and virtual (erect) images depending on the position of the object.

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A concave lens always forms a virtual, erect, and diminished (smaller) image, regardless of the object's distance.

📐Formulae

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

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

m=hiho=vum = \frac{h_i}{h_o} = \frac{v}{u}

💡Examples

Problem 1:

If the radius of curvature (RR) of a spherical lens is 30 cm30\text{ cm}, calculate its focal length (ff).

Solution:

f=302=15 cmf = \frac{30}{2} = 15\text{ cm}

Explanation:

The focal length of a lens is half of its radius of curvature. Using the formula f=R2f = \frac{R}{2}, we divide 3030 by 22 to get 15 cm15\text{ cm}.

Problem 2:

Calculate the magnification (mm) if the height of the object (hoh_o) is 5 cm5\text{ cm} and the height of the image (hih_i) formed by a lens is 10 cm10\text{ cm}.

Solution:

m=105=2m = \frac{10}{5} = 2

Explanation:

Magnification is the ratio of the height of the image to the height of the object. Since hi=10h_i = 10 and ho=5h_o = 5, the magnification is 22, meaning the image is twice as large as the object.

Problem 3:

Find the difference in focal lengths of two lenses, Lens A and Lens B, where Lens A has a focal length of 25 cm25\text{ cm} and Lens B has a focal length of 12 cm12\text{ cm}.

Solution:

25−1213\begin{array}{r} 25 \\ - 12 \\ \hline 13 \end{array}

Explanation:

Subtracting the focal length of Lens B from Lens A gives a difference of 13 cm13\text{ cm}.