the image of a mountain can be seen on the surface of calm water. what property of light is demonstrated?

Physics

2 answers:

raketka [301]3 years ago

6 0

Answer:

It is Reflection of light from water surface

Explanation:

When light coming from the mountains incident on the stationary water surface then major proportion of light gets reflected from the surface of water and this will received by the observer eye.

We also know that when light beam gets reflected into the form of beam then it is known as regular reflection.

Here when water surface is calm then the surface is assumed to be plane and in this case the light gets reflected by the surface of water in same direction and we got a good image in the water.

So this property of image is due to reflection of light.

Valentin [98]3 years ago

3 0

<span>I say it's the regular reflection</span>

You might be interested in

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$