Ask Your Teacher A piece of wire is bent through an angle θ. The bent wire is partially submerged in carbon disulfide (index of refraction n = 1.628) so that, to a person looking along the dry part, the wire appears to be straight and makes an angle of 26.4° with the horizontal. Determine the value of θ.

θ1 = 33.38°

Explanation:

From Snell's law of refraction,

n1•sinθ1 = n2•sinθ2

We are given ;

Angle of refraction; θ2 = (90 - 26.4°) = 63.6°

Index of refraction; n1 = 1.628

Index of refraction in air; n2 will have a constant value of 1

Now, we are looking for angle of incidence θ1, so let's make θ1 the subject;

n1•sinθ1 = n2•sinθ2

So, sinθ1 = (n2•sinθ2)/n1

Plugging in the relevant values;

sinθ1 = (1 x sin 63.6)/1.628

sinθ1 = 0.5502

θ1 = sin^(-1)0.5502

θ1 = 33.38°

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j