A light ray is incident at an angle 45° with the normal to a √2 cm thick plate (μ = 2). Find the shift in the path (in cm) of the light as it emerges out from the plate.

Question

A light ray is incident at an angle 45° with the normal to a √2 cm thick plate (μ = 2). Find the shift in the path (in cm) of the light as it emerges out from the plate. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

By Snell's law (sin i/sin r) = μ ∴ sin r = (sin i/μ) ...(i) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/abe9c53c4a68437aa9d3130528070a3b-.png" /> In Δ ABC, AC = (AB/cos r) In Δ ACD, CD = AC sin(i - r) = (AB/cos r)sin(i - r) ...(ii) On substituting AB = √2 cm, i = 45° and solving CD = 0.62 cm

Q. A light ray is incident at an angle $4 5^{\circ}$ with the normal to a $2 c m$ thick plate $(μ = 2)$ . Find the shift in the path (in cm) of the light as it emerges out from the plate.

By Snell's law
$\frac{s in i}{s in r} = μ$
$∴ s in r = \frac{s in i}{μ} ... (i)$

In $Δ A BC, A C = \frac{A B}{cos r}$
In $Δ A C D, C D = A C s in (i - r)$
$= \frac{A B}{cos r} s in (i - r) ... (ii)$
On substituting $A B = 2 c m, i = 4 5^{\circ}$
and solving $C D = 0.62 c m$

Q. A light ray is incident at an angle 45∘ with the normal to a 2​cm thick plate (μ=2). Find the shift in the path (in cm) of the light as it emerges out from the plate.

By Snell's law sinrsini​=μ ∴sinr=μsini​...(i) In ΔABC,AC=cosrAB​ In ΔACD,CD=ACsin(i−r) =cosrAB​sin(i−r)...(ii) On substituting AB=2​cm,i=45∘ and solving CD=0.62cm

Q. A light ray is incident at an angle $4 5^{\circ}$ with the normal to a $2 c m$ thick plate $(μ = 2)$ . Find the shift in the path (in cm) of the light as it emerges out from the plate.

By Snell's law
$\frac{s in i}{s in r} = μ$
$∴ s in r = \frac{s in i}{μ} ... (i)$

In $Δ A BC, A C = \frac{A B}{cos r}$
In $Δ A C D, C D = A C s in (i - r)$
$= \frac{A B}{cos r} s in (i - r) ... (ii)$
On substituting $A B = 2 c m, i = 4 5^{\circ}$
and solving $C D = 0.62 c m$