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Light: Shadows and Reflections - Pinhole Camera

Grade 7CBSE

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

🔑Concepts

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A pinhole camera works on the principle that light travels in a straight line, known as the rectilinear propagation of light.

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The image formed by a pinhole camera is always inverted (upside down) because the light rays from the top of the object and the bottom of the object cross each other at the pinhole.

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The image formed is a real image because it is captured on a screen (like tracing paper).

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The image is the same color as the object, which distinguishes it from a shadow.

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The size of the image depends on the ratio of the distance of the screen from the pinhole to the distance of the object from the pinhole.

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If the distance between the pinhole and the screen (vv) increases, the size of the image (hih_i) increases.

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If the distance between the object and the pinhole (uu) increases, the size of the image (hih_i) decreases.

📐Formulae

Magnification (m)=Height of Image (hi)Height of Object (ho)\text{Magnification } (m) = \frac{\text{Height of Image } (h_i)}{\text{Height of Object } (h_o)}

hiho=vu\frac{h_i}{h_o} = \frac{v}{u}

Height of Image (hi)=ho×vu\text{Height of Image } (h_i) = \frac{h_o \times v}{u}

💡Examples

Problem 1:

An object of height ho=50 cmh_o = 50\text{ cm} is placed at a distance u=200 cmu = 200\text{ cm} from a pinhole camera. If the length of the camera (distance from pinhole to screen) is v=20 cmv = 20\text{ cm}, calculate the height of the image hih_i.

Solution:

Using the formula hi=ho×vuh_i = \frac{h_o \times v}{u}, we substitute the values: hi=50×20200h_i = \frac{50 \times 20}{200} hi=1000200=5 cmh_i = \frac{1000}{200} = 5\text{ cm}

Explanation:

The light rays from the 50 cm50\text{ cm} object converge at the pinhole and diverge to form a 5 cm5\text{ cm} inverted image on the screen because the screen is 1010 times closer to the pinhole than the object is.

Problem 2:

A student observes two images of a candle. The first image height is 12 cm12\text{ cm} and the second image height is 8 cm8\text{ cm} after moving the camera further away. Calculate the difference in the heights of the two images.

Solution:

To find the difference, we subtract the smaller height from the larger height: 12−84\begin{array}{r} 12 \\ - 8 \\ \hline 4 \end{array} The difference is 4 cm4\text{ cm}.

Explanation:

Moving the camera further from the object (increasing uu) or moving the screen closer to the pinhole (decreasing vv) results in a smaller image. The vertical calculation shows the reduction in size was 4 cm4\text{ cm}.