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The Human Eye and Optical Phenomena - shows the actual and apparent positions of the

Grade 10CBSE

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

🔑Concepts

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Atmospheric Refraction: The refraction of light caused by the Earth's atmosphere due to variations in the optical density of air layers. Air layers closer to the Earth are denser than layers higher up.

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Apparent Position of Stars: As starlight enters the Earth's atmosphere, it undergoes continuous refraction. Since the refractive index of air increases towards the surface, the light bends towards the normal. Therefore, the star appears slightly higher than its actual position.

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Advanced Sunrise and Delayed Sunset: The Sun is visible to us about 22 minutes before the actual sunrise and 22 minutes after the actual sunset. This is because light from the Sun, when it is below the horizon, is refracted by the atmosphere towards the observer.

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Apparent Depth: When an object is placed in a denser medium (like water or glass) and viewed from a rarer medium (like air), it appears to be at a lesser depth than it actually is. The ratio of real depth to apparent depth is equal to the refractive index nn of the medium.

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Twinkling of Stars: The physical conditions of the Earth's atmosphere change continuously, causing the apparent position of the star and the amount of starlight reaching the eye to fluctuate. This causes the star to 'twinkle'.

📐Formulae

n=Real DepthApparent Depthn = \frac{\text{Real Depth}}{\text{Apparent Depth}}

Apparent Shift=Real Depth−Apparent Depth\text{Apparent Shift} = \text{Real Depth} - \text{Apparent Depth}

Apparent Shift=t(1−1n) (where t is the real thickness or depth)\text{Apparent Shift} = t \left( 1 - \frac{1}{n} \right) \text{ (where } t \text{ is the real thickness or depth)}

Total difference in day length=2 minutes (sunrise)+2 minutes (sunset)=4 minutes\text{Total difference in day length} = 2 \text{ minutes (sunrise)} + 2 \text{ minutes (sunset)} = 4 \text{ minutes}

💡Examples

Problem 1:

A swimming pool appears to be 3 m3 \text{ m} deep when viewed from above. If the refractive index of water is n=1.33n = 1.33 (or 43\frac{4}{3}), find the actual depth of the pool and the vertical shift of the bottom.

Solution:

Given: Apparent Depth =3 m= 3 \text{ m}, n=43n = \frac{4}{3}. Using the formula n=Real DepthApparent Depthn = \frac{\text{Real Depth}}{\text{Apparent Depth}}, we have: 43=Real Depth3\frac{4}{3} = \frac{\text{Real Depth}}{3} Real Depth=4 m\text{Real Depth} = 4 \text{ m}. To find the shift: 4−31\begin{array}{r} 4 \\ -3 \\ \hline 1 \end{array} The shift is 1 m1 \text{ m}.

Explanation:

The pool appears shallower because light rays traveling from the denser medium (water) to the rarer medium (air) bend away from the normal, making the bottom of the pool appear raised.

Problem 2:

Explain why the Sun appears flattened at sunrise and sunset.

Solution:

The lower edge of the Sun's disc is closer to the horizon than the upper edge. Thus, light from the lower edge travels through more atmosphere and undergoes more refraction than light from the upper edge. This difference in refraction results in the apparent flattening of the Sun's disc.

Explanation:

This is a direct consequence of atmospheric refraction affecting different parts of the large solar disc by slightly different magnitudes.