Physics Ray Optics and Optical Instruments
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New answer posted
a month agoContributor-Level 10
Whenever a light ray travels from a rarer medium to a denser medium through a spherical surface, relationship between object distance (u), image distance (v), radius of curvature (R) and refractive index ( and ) is given by refraction formula for spherical surfaces as follows.
- is the refractive index of medium that contains the object (on the left side)
- is refractive index of medium that will contain the image (on the right side)
New answer posted
a month agoContributor-Level 10
Yes, total internal reflection can occur in both sound and water waves, including light waves, since it follows the same principle. Whenever a light/sound/water wave travels from denser to rarer medium and strikes the boundary of the medium at an angle greater than critical angle, the wave entirely reflects back into denser medium rather than refracting. Sonar waves in submarine reflect back entirely from the water surface as it hits at a steep angle. In water, ocean waves approach beach and slow down because of shallower water. If a wave hits the boundary of deep and shallow water at a steep angle, these waves will reflect back toward
New answer posted
a month agoContributor-Level 10
Let us take a look at the various applications of Total Internal Reflection in real life:
Total internal reflection is the foundational phenomenon used in modern telecommunications, internet cables and medical imaging. Light travels through thin and flexible fibres, which are made of glass and plastic. This light, then, undergoes Total internal reflection in the inner walls of the fibre, which allows travelling long distances with minimum loss.
Mirage is a phenomenon that occurs due to the total internal reflection as light gets reflected between the layers of air.
Diamonds shine because of total internal reflection. The gem is cut
New answer posted
a month agoContributor-Level 10
Mirage is formed due to total internal reflection in the following way:
On a hot day, the air near the ground becomes extremely hot as compared to air above it since hot air is less dense than cooler air, which has a lower refractive index.
Light from the sky travels toward hotter and less dense air near the ground. As this light enters the hotter layer, it will bend away from the normal because of the change in refractive index.
In case the angle of light is steep enough, it will exceed the critical angle between two layers of air. The light, instead of bending, reflects back up as a result of total internal reflection.
This reflected lig
New answer posted
a month agoContributor-Level 10
θ? > I = θ ⇒ sinθ? > sinθ ⇒ 1/μ > sinθ
⇒ μ < 1/sin
Note: If we assume θ = 45° ⇒ μ < 1.414, then red colour light ray will come out from face PR of prism.
New answer posted
a month agoContributor-Level 10
Given the refractive index μ = λ? / λ = 3/2 and v = 10m, the object distance is u = - (3/2)v = -15m.
Using the lens maker's formula:
μ/v - 1/u = (μ - 1)/R
(3/2)/10 - 1/ (-15) = (3/2 - 1)/R
This gives the radius of curvature R:
R = -30/13 m
New question posted
a month agoNew answer posted
2 months agoContributor-Level 10
Image by objective must be at focus of eye piece.
1/v - 1/u = 1/f?
1/5 - 1/u = 1 ; u = -5/4 cm
Hence, N = 50.
New answer posted
2 months agoContributor-Level 10
V = I? (G + R? )
1 = I? (G + R? )
2 = I? (G + R? + R? )
2 = 1 + I? R?
I? R? = 1
R? = G + R?
New answer posted
2 months agoContributor-Level 10
Refractive index (n) is a dimensionless quantity that has no units. Since both numerator and denominator are in meters per seconds (m/s), both units cancel each other. This makes refractive index simply a number.
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