If the angle of reflection of a propagated ray is 35°. Calculate the a. Angle of incidence b. Angle of deviation c. Angle of glance

First of all,

Angle of incidence = Angle between normal and incident ray.Angle of reflection = Angle between normal and reflected rayGlance angle of incidence = Angle between incident ray and the surface.Glance angle of reflection = Angle between reflected ray and the surface.

We have,

➝ Angle of reflection = 35°

(a) According to law of reflection,

Angle of incidence = Angle of reflection

➝ Angle of incidence = 35°

(b) By using formula,

Angle of Deviation = 180° - 2(Angle of incidence)

From (a),

➝ Angle of Deviation = 180° - 2(35°)

➝ Angle of Deviation = 180° - 70°

➝ Angle of Deviation = 110°

(c) We know,

Angle of Incidence + Glance angle of incidence = 90°

Angle of reflection + Glance angle of reflection = 90°

So,

As angle of incidence = angle of reflection

Then, Angle of glance is also equal.

➝ Angle of glance = 90° - 35° = 55°

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Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge