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Light: Shadows and Reflections - Making Some Useful Items

Grade 7CBSE

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

🔑Concepts

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Rectilinear Propagation of Light: Light travels in a straight line. This property is the fundamental principle behind the formation of shadows and the working of a pinhole camera.

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Pinhole Camera: A simple device without a lens that forms an inverted image of an object on a translucent screen. The image formed is a real image because it can be obtained on a screen.

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Characteristics of Pinhole Images: The image is always inverted (upside down), shows the colors of the object, and its size varies depending on the distance between the pinhole and the screen (vv) and the distance between the pinhole and the object (uu).

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Reflection of Light: The process of bouncing back of light rays when they fall on a polished or shiny surface like a mirror. A plane mirror changes the direction of light that falls on it.

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Periscope: An optical instrument that uses two plane mirrors placed parallel to each other at an angle of 45∘45^{\circ} to the line of sight. It allows an observer to see objects that are not in the direct line of sight (e.g., seeing over a wall or from a submarine).

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Shadows vs. Images: A shadow is always dark and only shows the outline of the object, whereas an image formed by a pinhole camera or mirror shows the color, details, and shape of the object.

📐Formulae

Law of Reflection: ∠i=∠r\text{Law of Reflection: } \angle i = \angle r

Magnification in Pinhole Camera (m)=Height of Image (hi)Height of Object (ho)=Distance of Screen from Pinhole (v)Distance of Object from Pinhole (u)\text{Magnification in Pinhole Camera } (m) = \frac{\text{Height of Image } (h_i)}{\text{Height of Object } (h_o)} = \frac{\text{Distance of Screen from Pinhole } (v)}{\text{Distance of Object from Pinhole } (u)}

💡Examples

Problem 1:

An object of height 15 cm15\text{ cm} is placed at a distance of 100 cm100\text{ cm} from a pinhole camera. If the distance between the pinhole and the screen is 20 cm20\text{ cm}, calculate the height of the image formed.

Solution:

Given: ho=15 cmh_o = 15\text{ cm}, u=100 cmu = 100\text{ cm}, v=20 cmv = 20\text{ cm}. Using the formula hiho=vu\frac{h_i}{h_o} = \frac{v}{u}, we get hi=v×hou=20×15100=300100=3 cmh_i = \frac{v \times h_o}{u} = \frac{20 \times 15}{100} = \frac{300}{100} = 3\text{ cm}.

Explanation:

The height of the image formed on the screen is 3 cm3\text{ cm}. The image will be inverted and smaller than the object.

Problem 2:

In a periscope, if the first mirror reflects light at an angle, what must be the orientation of the second mirror to allow the light to reach the observer's eye parallel to the original path?

Solution:

The two mirrors must be placed parallel to each other, both making an angle of 45∘45^{\circ} with the frame of the periscope.

Explanation:

Light hits the first mirror and turns 90∘90^{\circ}. It then hits the second mirror and turns another 90∘90^{\circ}, making it parallel to the original entry path but at a different height.