Answer:
(a). The thickness of the glass is 868 nm.
(b). The wavelength is 3472 nm.
Explanation:
Given that,
Refractive index = 1.20
Wavelength = 496 nm
Next wavelength = 386 nm
We need to calculate the thickness of the glass
Using formula for constructive interference
Put the value into the formula
In first case,
.....(I)
In second case,
.....(II)
From equation (I) and (II)
Put the value of m in equation (I)
The thickness of the glass is 868 nm.
(b). We need to calculate the wavelength
Using formula of constructive interference
Put the value into the formula
Hence, (a). The thickness of the glass is 868 nm.
(b). The wavelength is 3472 nm.
Answer:
a) 120 N
b) 5 N
c) 0.2 N
d) Mass remains the same, and weight decreases.
Explanation:
<em>Use the formula W = mg, where mass is in kg, and gravitational field strength in N/kg.</em>
a)
W = mg
= 12 × 10
= <u>120 N</u>
b)
500 g = 0.5 kg
W = mg
= 0.5 × 10
= <u>5 N</u>
c)
20 g = 0.02 kg
W = mg
= 0.02 × 10
= <u>0.2 N</u>
d)
<u>Mass remains the same, and weight decreases.</u>
Answer:
Aluminum has the highest atomic number
Answer:
0.5639m
Explanation:
For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ
=sin⁻¹(mλ/d)
mλ /d =y/L
for the first order,
y= mλL/d
For ratio separation y₀/yD=1 and d= 1
y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]
1=λ ₀L₀/λD.LD.
λD.LD= λ ₀L₀
L₀= λD.LD/ λ ₀..............(1)
Then substitute the given values into (1) we have
L₀=471 *0.497/611
= 0.3831m
Distance by which the screen has to be moved towards the slit is
LD- Lo
0.947-0.3831= 0.5639m