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This is one of our many quizzes on Refraction. Click on this to access more such competitive quizzes on Ray Optics.

Refraction Quiz 2

physics

 The angle of refraction for the given ray diagram is ____ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/26b7d4ee-a642-4864-9288-2779050e795e7233290260992923484.png"/> 

  According to Snell’s law, the angle of refraction is equal to the angle of incidence times the ratio of refractive indices of the mediums. 

  $\text{According to Snell }\!\!'\!\!\text{ s law}\text{.} \\ \Rightarrow \dfrac{\sin \,i}{\sin \,r}=\dfrac{{{n}_{2}}}{{{n}_{1}}} \\ \text{Substitute the values in the above formula}\text{. } \\ \Rightarrow \dfrac{\sin \,45{}^\circ }{\sin \,r}=\dfrac{1.33}{1} \\ \Rightarrow \sin r=\dfrac{{}^{1}/{}_{\sqrt{2}}}{1.33}=0.53 \\ \therefore r=32{}^\circ $

  The angle of refraction for the given ray diagram is ____ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/705b83ee-4b56-4d8b-a022-746d45d297bd4194986538080608291.png"/>

 The sum of the incident angle and the glancing angle equals $\text{90}{}^\circ $. For the same value of the refractive index, the angle of incidence equals the angle of refraction.  

  The difference between the glancing angle and the incident angle is ${90}{}^\circ$. $ \\ \text{i = 90}{}^\circ \text{- g} \\ $ Substitute the value of glancing angle in the above equation. $ \\ \Rightarrow \text{i = 90}{}^\circ \text{- }30{}^\circ  \\ \therefore \text{i = 60}{}^\circ  \\ $ For the same refractive index, the angle of incidence equals the angle of refraction. $ \\ \Rightarrow \text{sin i = sin r} \\ \Rightarrow \text{r = i} \\ \therefore \text{r = 60}{}^\circ $

  The angle of incidence and refraction for the angle of minimum deviation $30{}^\circ $ is ___ and ___? $[$Take $A=60{}^\circ ]$

 The angle of deviation equals the difference between sum of first and second incidence angles and angle of prism. The angle of refraction equals half of the angle of prism at minimum deviation.  At minimum deviation, both first and second angles of incidence and refraction will be equal.

 At minimum deviation, $i=i^\prime$ and $r=r^\prime$, The angle of deviation equals to, $ \\ \delta =i+i^\prime -A \\ \Rightarrow \delta =2i-A \\ $ Substitute the values in the above formula. $ \\ \Rightarrow 30{}^\circ +60{}^\circ =2i \\ \Rightarrow i=\dfrac{90{}^\circ }{2} \\ \therefore i=45{}^\circ  \\ $ The angle of refraction at minimum deviation equals to, $ \\ r+r^\prime =A \\ \Rightarrow 2r=A \\ $ Substitute the values in the above formula. $ \\ \Rightarrow 2r=60{}^\circ  \\ \therefore r=30{}^\circ $

  For $e=\dfrac{3}{4}A$, compute the angle of refraction. 

 The sum of the angle of prism and the angle of deviation equals the sum of angle of incidence and the angle of emergence. The angle of refraction equals the difference between the angles of incidence and deviation.  

 The relation between the angles of prism, deviation, incidence and emergence is,  $ \\ \Rightarrow A+\delta =i+e \\ \Rightarrow \delta =2i-A \\ $ Substitute the values in the above formula. $ \\ \Rightarrow \delta =2\times \dfrac{3}{4}A-A \\ \Rightarrow \delta =\dfrac{1}{2}A=\dfrac{1}{2}\times 60{}^\circ  \\ \therefore \delta =30{}^\circ  \\ $ The angle of refraction equals the difference between the angles of incidence and deviation. $ r=i-\delta $  $ \\ $ Substitute the values in the above formula. $ \\ \Rightarrow r=\dfrac{3}{4}\times 60{}^\circ -30{}^\circ  \\ \Rightarrow r=45{}^\circ -30{}^\circ  \\ \therefore r=15{}^\circ $

 Compute the refractive index of prism for the angle of deviation $10{}^\circ $. The angle of refraction for the angle of incidence $60{}^\circ $ is ____

 Use the refractive index of the prism formula. The angle of incidence equals half the sum of angles of prism and deviation. The angle of refraction equals the angle of incidence by the refractive index. 

 The refractive index of the prism is, $ \\ \Rightarrow \dfrac{{{n}_{prism}}}{{{n}_{0}}}=\dfrac{\sin \left( \dfrac{A+\delta }{2} \right)}{\sin \left( \dfrac{A}{2} \right)} \\ $ Substitute the values in the above formula. $ \\ \Rightarrow \dfrac{{{n}_{prism}}}{1}=\dfrac{\sin \left( \dfrac{60{}^\circ +10{}^\circ }{2} \right)}{\sin \left( \dfrac{60{}^\circ }{2} \right)} \\ \Rightarrow {{n}_{prism}}=\dfrac{\sin \left( 35{}^\circ  \right)}{\sin \left( 30{}^\circ  \right)} \\ \therefore {{n}_{prism}}=1.147 \\ $ The angle of incidence is, $ \\ \delta =2i-A \\ \Rightarrow i=\dfrac{A+\delta }{2} \\ $ Substitute the values in the above formula. $ \\ i=\dfrac{60{}^\circ +10{}^\circ }{2} \\ \therefore i=35{}^\circ  \\ $ According to Snell’s law, $ \\ \dfrac{{{n}_{prism}}}{{{n}_{0}}}=\dfrac{\sin i}{\sin r} \\ $ Substitute the values in the above formula. $ \\ \Rightarrow \dfrac{1.147}{1}=\dfrac{\sin 35{}^\circ }{\sin r} \\ \Rightarrow \sin r=\dfrac{\sin 35{}^\circ }{1.147} \\ \therefore r=30{}^\circ $