angle deviated through prism & minimum deviation

Question

A beam of white light makes an angle of incidence of 38° with a glass prism of index of refraction of 1.5 and refracting angle of 60°. Through what angle is the ray deviated by the prism? Does this represent minimum deviation?

Answer 1

Step 1: Refraction at first face

sin⁡r1=sin⁡i1n=sin⁡38∘1.5⇒r1≈24.23∘\sin r_1=\frac{\sin i_1}{n}=\frac{\sin 38^\circ}{1.5}\Rightarrow r_1\approx 24.23^\circsinr1​=nsini1​​=1.5sin38∘​⇒r1​≈24.23∘Step 2: Inside the prism

r1+r2=A⇒r2=60∘−24.23∘≈35.77∘r_1+r_2=A \Rightarrow r_2=60^\circ-24.23^\circ\approx 35.77^\circr1​+r2​=A⇒r2​=60∘−24.23∘≈35.77∘Step 3: Emergence at second face

sin⁡i2′=nsin⁡r2=1.5sin⁡35.77∘⇒i2′≈61.25∘\sin i_2' = n\sin r_2 = 1.5\sin 35.77^\circ \Rightarrow i_2'\approx 61.25^\circsini2′​=nsinr2​=1.5sin35.77∘⇒i2′​≈61.25∘Deviation

δ=i1+i2′−A=38∘+61.25∘−60∘≈39.3∘\delta = i_1 + i_2' - A = 38^\circ + 61.25^\circ - 60^\circ \approx \boxed{39.3^\circ}δ=i1​+i2′​−A=38∘+61.25∘−60∘≈39.3∘​Minimum deviation check

For minimum deviation, r1=r2=A/2=30∘r_1=r_2=A/2=30^\circr1​=r2​=A/2=30∘ and

δmin⁡=2arcsin⁡ ⁣(nsin⁡A2)−A=2arcsin⁡(1.5⋅12)−60∘≈37.18∘,\delta_{\min}=2\arcsin\!\big(n\sin\frac{A}{2}\big)-A =2\arcsin(1.5\cdot \tfrac{1}{2})-60^\circ \approx 37.18^\circ,δmin​=2arcsin(nsin2A​)−A=2arcsin(1.5⋅21​)−60∘≈37.18∘,with corresponding imin⁡=arcsin⁡(nsin⁡30∘)=arcsin⁡(0.75)≈48.6∘i_{\min}=\arcsin(n\sin 30^\circ)=\arcsin(0.75)\approx 48.6^\circimin​=arcsin(nsin30∘)=arcsin(0.75)≈48.6∘.

Since the given incidence 38∘≠48.6∘38^\circ \ne 48.6^\circ38∘=48.6∘ and δ≈39.3∘>δmin⁡≈37.2∘\delta\approx 39.3^\circ > \delta_{\min}\approx 37.2^\circδ≈39.3∘>δmin​≈37.2∘, this is not minimum deviation.