Question
Question: In YDSE, find the thickness of a glass slab (in\( \times \) \({\text{1}}{{\text{0}}^{\text{4}}}\math...
In YDSE, find the thickness of a glass slab (in× 104A0) (μ = 1.5) which should be placed before the upper slit S1 so that central maximum now lies at a point where 5th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is 5000A0.
Solution
To solve this problem, we should calculate the lateral shift. If the values of lateral displacement, angle of incidence and angle of refraction are known then the thickness of the glass slab can be determined.
Formula used: The formula used here, would be:
Lateral shift: X - = λβ(μ - 1)t = nβ
Here, nis the wavelength.
μis the refractive index.
n is the number of fringes.
Complete step-by-step answer:
Let us assume that, initially the light falls and it has a wavelength of 5000 A0.
Applying the formula,
Lateral shift: X - = λβ(μ - 1)t = nβ
t = β(μ - 1)5βλ
β cancels on both numerator and denominator,
And, on putting the values,
⇒(1.5 - 1)5×5000
=50000 A0
Now, we need the thickness of slab in × 104A0,
So, t = 5 or 5 ×104A0.
Additional Information:
Young's Double slit experiment, when it was first carried out by Young in 1801, demonstrated that light was a wave, which seemed to clear the confusion whether light was corpuscular (particle-like) or wave. When a ray of light is incident obliquely on a parallel sided glass slab the emergent ray shifts laterally. The perpendicular distance between the direction of the incident ray and emergent ray is called lateral shift.
Also, the apparent shift in the position of an object placed in one medium and viewed along the normal, from the other medium is known as normal shift.
Note:
The mistake that could be done here is in taking the units as all the mathematical functions can be performed in the same unit.
Some other applications of YDSE could be to find the position of dark fringes or the position of light fringes, fringe width etc.