what is understood by lateral displacement of light? illustrate it with the help of a diagram. list any two factors on which the lateral displacement of a perfectly kill a substance depends.
❌don't spam ❌
Answers
When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges parallel to its incident direction. Suppose a light ray enters the glass slab from air. It will first be refracted at this surface. ... The distance between these two rays is called the lateral displacement.
Answer:
μ1sini=μ2sinr , where i is the angle of incidence and r is the angle of refraction at the interface of the two mediums. μ1 is the refractive index of the medium where the light is incident and μ2 is the refractive index of the medium where the light gets refracted.
Complete step-by-step solution -
A glass slab is in cuboidal shape (opposite surfaces parallel to each other). Let us first understand what happens when a light ray enters a glass slab. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges parallel to its incident direction. Suppose a light ray enters the glass slab from air. It will first be refracted at this surface.
Diagram above↑
Use Snell’s law i.e. μ1sini=μ2sinr.
⇒1.sini=(1.5).sinr …….(1) (Refractive indices of air and glass are 1 and 1.5 respectively)
Afterwards, the refracted ray will travel through the glass and will be incident on the opposite surface. Here, as you can see in the given figure, the angle of incident at surface 2 will be equal to the angle of refraction at surface 1 (since the normal at the two surfaces are parallel).
Let the angle of refraction at surface two be ‘e’. Now, use Snell’s law.
⇒1.sine=(1.5)sinr …….(2)
Compare equations (1) and (2). We get, i=e. Hence, the emergent ray is parallel to the incident ray. The distance between these two rays is called the lateral displacement. Now, let us find an expression for the lateral displacement. For that, let the thickness of the glass slab be ‘t’.
As you can see in the given figure, △OMN is a right-angled triangle where MN is the lateral displacement. We have an expression for MN. In △OTM , sinδ=MNOM
⇒MN=OMsinδ ….(3).
In △OTM , cosr=OTOM=tOM
⇒OM=tsecr ……(4).
Substitute the value of OM from equation (4) into equation (3).
⇒MN=tsecr.sinδ And δ=i−r
⇒MN=tsecr.sin(i−r)
Therefore, we got the expression for lateral displacement. Now, if the light ray is incident normally on the surface, i=0. Hence, r=0. Therefore, MN will be zero. Meaning there will be any displacement in the ray of light.
Note: When a light ray falls on a glass slab, it is not that there will be a lateral displacement every time. One example you saw in the question. Another example is the internal reflection of the ray. The ray does refract into the outside medium, but it is reflected back into the same medium. This is only when the ray travels from a denser medium to a rarer medium.
Kkrh??
good evening
