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Light - Shadows and Eclipses

Grade 6ICSE

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

🔑Concepts

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Rectilinear Propagation of Light: Light always travels in a straight line. This property is responsible for the formation of shadows and the working of a pinhole camera.

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Shadow: A dark region formed when an opaque object blocks the path of light. A shadow consists of two parts: the Umbra (the region of total darkness) and the Penumbra (the region of partial darkness).

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Opaque, Transparent, and Translucent: Transparent objects allow all light to pass through; Translucent objects allow partial light; Opaque objects block light completely, creating sharp shadows.

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Pinhole Camera: A simple device without a lens that forms a real, inverted image of an object on a screen. The image size changes based on the distance of the object from the pinhole.

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Solar Eclipse: Occurs on a New Moon day when the Moon passes between the Sun and the Earth, casting its shadow on the Earth. The Sun's rays are blocked by the Moon.

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Lunar Eclipse: Occurs on a Full Moon night when the Earth passes between the Sun and the Moon, casting its shadow on the Moon. The Moon enters the Earth's Umbra or Penumbra.

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Speed of Light: Light travels at a very high speed, approximately 3imes108extm/s3 imes 10^8 ext{ m/s} in a vacuum or air.

📐Formulae

Speed of light (c)≈3×108 m/s\text{Speed of light (c)} \approx 3 \times 10^8 \text{ m/s}

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} where vv is the distance of the screen from the pinhole and uu is the distance of the object from the pinhole.

💡Examples

Problem 1:

An object of height 15extcm15 ext{ cm} is placed at a distance of 60extcm60 ext{ cm} from a pinhole camera. If the length of the camera box (distance to screen) is 20extcm20 ext{ cm}, calculate the height of the image formed.

Solution:

Given: ho=15extcmh_o = 15 ext{ cm}, u=60extcmu = 60 ext{ cm}, v=20extcmv = 20 ext{ cm}. Using the formula hiho=vu\frac{h_i}{h_o} = \frac{v}{u}, we get hi=ho×vu=15×2060=30060=5extcmh_i = \frac{h_o \times v}{u} = \frac{15 \times 20}{60} = \frac{300}{60} = 5 ext{ cm}.

Explanation:

The image height is 5extcm5 ext{ cm}, and it will be inverted because of the rectilinear propagation of light through the pinhole.

Problem 2:

During a Solar Eclipse, which celestial body is in the middle, and what phase of the moon is it?

Solution:

The Moon is in the middle (between the Sun and the Earth). This occurs during the New Moon phase.

Explanation:

For a shadow to fall on Earth, the Moon must be positioned such that it blocks the Sun's light, which only happens during the New Moon alignment.

Shadows and Eclipses - Revision Notes & Key Formulas | ICSE Class 6 Science