1.A mica strip and a polysterene strip are fitted on the two slits of YDSE apparatus. The thickness of strip is 0.5mm, distance betwn slits=0.12cm. The R.I of mica and polysterene is 1.58 and 1.55 respectively. wavelength of light used is 590nm. At what distance from centre will the first maximum be located?

2. A glass surface is coated by an oil film of uniform thickness 10^-4 cm. The R.I of oil is 1.25 and that of glass is 1.5. Find the wavelengths of light in visible region(400nm-750nm) which are completely transmitted by oil film under normal incidence.

To tackle these two questions, we need to apply concepts from wave optics, particularly related to Young's Double Slit Experiment (YDSE) and thin film interference. Let’s break down each problem step by step.

Finding the Position of the First Maximum in YDSE

In the first scenario, we have a Young's Double Slit Experiment setup where two different materials, mica and polystyrene, are placed in front of the slits. The thickness of these strips and their refractive indices will affect the path difference of the light waves passing through them.

Step 1: Calculate the Optical Path Difference

The optical path difference (OPD) introduced by a material can be calculated using the formula:

OPD = (n - 1) * d

where n is the refractive index of the material and d is the thickness of the material. For the mica strip:

• Thickness, d = 0.5 mm = 0.05 cm
• Refractive Index of Mica, n_mica = 1.58
• OPD_mica = (1.58 - 1) * 0.05 = 0.029 cm

For the polystyrene strip:

• Refractive Index of Polystyrene, n_polystyrene = 1.55
• OPD_polystyrene = (1.55 - 1) * 0.05 = 0.0275 cm

Step 2: Total Path Difference

The total path difference (Δ) between the two slits due to the two materials is:

Δ = OPD_mica - OPD_polystyrene = 0.029 cm - 0.0275 cm = 0.0015 cm

Step 3: Calculate the Position of the First Maximum

The position of the m-th maximum in YDSE is given by:

y_m = (m * λ * D) / d

Where:

• y_m = position of the m-th maximum
• m = order of the maximum (for the first maximum, m = 1)
• λ = wavelength of light = 590 nm = 0.000059 cm
• D = distance from the slits to the screen (not given, but we can assume it is large enough)
• d = distance between the slits = 0.12 cm

Substituting m = 1, we find:

y_1 = (1 * 0.000059 * D) / 0.12

Now, we need to consider the path difference due to the optical path difference:

y_1 = (1 * 0.000059 * D) / 0.12 + Δ

Since we don’t have the value of D, we can express the position of the first maximum in terms of D:

y_1 = (0.000059 * D) / 0.12 + 0.0015

Wavelengths Transmitted by Oil Film on Glass

Now, let’s move on to the second problem regarding the oil film on a glass surface. Here, we need to find the wavelengths of light that are completely transmitted through the oil film.

Step 1: Determine the Condition for Constructive Interference

For a thin film, the condition for constructive interference (where light is transmitted) is given by:

2 * n_f * t = (m + 0.5) * λ

where:

• n_f = refractive index of the film (oil)
• t = thickness of the film
• m = order of interference (an integer)
• λ = wavelength of light in vacuum

Step 2: Substitute Known Values

Given:

• Thickness, t = 10^-4 cm = 10^-6 m
• Refractive Index of Oil, n_f = 1.25

Substituting these values into the equation gives:

2 * 1.25 * 10^-6 = (m + 0.5) * λ

This simplifies to:

2.5 * 10^-6 = (m + 0.5) * λ

Step 3: Finding Wavelengths in the Visible Range

Now, we can find the wavelengths for different values of m:

• For m = 0: λ = 2.5 * 10^-6 / 0.5 = 5 * 10^-6 m (5000 nm, not visible)
• For m = 1: λ = 2.5 * 10^-6 / 1.5 = 1.67 * 10^-6 m (1670 nm, not visible)
• For m = 2: λ = 2.5 * 10^-6 / 2.5 = 1 * 10^-6 m (1000 nm, not visible)
• For m = 3: λ = 2.5 * 10^-6 / 3.5 = 714 nm (visible)
• For m = 4: λ = 2.5 * 10^-6 / 4.5 = 555 nm (visible)
• For m = 5: λ = 2.