Answer:
Let the second medium be air (n₁=1)
The refractive index n₂ of the medium where first medium is air is found (a)
(a) n₂ = 2
Explanation:
Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.
Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.
Critical angle and Refractive index are related by:
To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)
Also θ(critical)=30°
Find n₂ :
Electrostatic repulsion is the force between two charges having the same sign, that tends to separate them further. The force is proportional to the product of the charges, and inversely proportional to the square of the distance between them.
We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
D = 1/f, where D is the power in diopters and f is the focal length in meters.
D=1/20
<u>D=0.05</u>
Answer:
Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).
