1. Coherent beams with wavelengths of 400 nm intersect at points on the screen. What will be observed at these points interference maximum or minimum, if the optical path difference of the beam is in the 1st case A1=20 um and in the 2nd - A2=15 um? Find the order of minimum and maximum. 2. An installation for observing Newton's rings in reflected light is illuminated by monochromatic light incident normally to the plate surface. The distance between the second and fourth dark rings is 5 mm. Find the distance between the eighth and sixteenth light rings. 3. Find the angle between the main planes of the polarizer and the analyzer if the intensity of the natural light passing through the polarizer and the analyzer decreased by 5 times.

Answers

Answer 1

Answer:

1. The observed interference at points will be maximum for $A_1 = 20 \, \mu m$ and minimum for $A_2 = 15 \, \mu m$ , with orders 12th maximum and 12th minimum.
2. The distance between the eighth and sixteenth light rings is approximately 2.583 mm.
3. The angle between the main planes of the polarizer and the analyzer is approximately $63.43^\circ$ when the intensity decreases by 5 times.

1. For coherentbeams with wavelengths of 400 nm, the interference observed at points on the screen will be maximum when the optical path difference (OPD) is a whole number of wavelengths ( $m \lambda$ ), and minimum when the OPD is a half-integer number of wavelengths ( $(m + 0.5) \lambda$ ).

Given

$A_1 = 20 \, \mu m$

$A_2 = 15 \, \mu m$ , the difference in optical path between the two cases is $$\Delta A = A_1 - A_2 = 5 \, \mu m$.$

The number of wavelengths difference can be calculated as $(\Delta A)/(\lambda)$ , which is approximately 12.5 wavelengths.

2. In Newton's ringspattern, the radius of the m dark ring is given by $r_m = √(m \lambda R)$ , where R is the radius of curvature of the lens.

Given the distance between the second and fourth darkrings ( $m = 2\) to \(m = 4$ ) as 5 mm, we can set up an equation:

$$√(4 \lambda R) - √(2 \lambda R) = 5 \, \text{mm}$.$

Solving for R gives $$R \approx 1.904 * 10^(-2) \, \text{m}$.$

Now, the radius of the 16th light ring can be calculated as $r_(16) = √(16 \lambda R)$ , and the radius of the 8th light ring is $r_(8) = √(8 \lambda R)$ . The distance between these two light rings is then given by:

$$d = r_(16) - r_(8) \n\n= √(16 \lambda R) - √(8 \lambda R)$.$

Plugging in the values, we get $d \approx 2.583 \, \text{mm}$ .

3. When the intensity decreases by a factor of 5, the transmission of the polarizer and analyzer combination becomes $(1)/(√(5))$ times the original transmission, as intensity is proportional to the square of the transmission.

The relationship between the intensity and the angle $\theta$ between the main planes of the polarizer and the analyzer is given by Malus's law: $I = I_0 \cos^2 \theta$ , where $I_0$ is the initial intensity.

Given that $I_0$ decreases by a factor of 5, we have:

$(I)/(I_0) = (1)/(√(5)) \n\n= \cos^2 \theta$

Solving for $\theta$ , we get:

$\theta = \arccos \left( (1)/(√(5)) \right)\n\n \approx 63.43^\circ$

Thus, the angle is $63.43^\circ$ .

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Answer 2

Answer:

Answer:

a=3....................................

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