MBUYE LIGHT NOTES PART 2.pdf

SSEKWE ROBERT 𝑨 𝒅 REFRACTION THROUGH PRISMS A prism has two refracting surfaces. It uses a principle of reversibility of light, therefore light can be incident on either surface. When a ray of light is incident on the first surface at an angle of incidence 𝑖1, it is refracted at an angle of refraction, 𝑟1 with an angle of deviation, 𝑑1. The ray then emerges out on the second surface at an angle of emergence 𝑖2, with an angle of refraction, 𝑟2 and angle of deviation, 𝑑2. 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝒅𝟏= 𝒊𝟏−𝒓𝟏 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝒅𝟐= 𝒊𝟐−𝒓𝟐 𝑇𝑜𝑡𝑎𝑙 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝒅= 𝒅𝟏+𝒅𝟐 𝒅= ( 𝒊𝟏−𝒓𝟏) + ( 𝒊𝟐−𝒓𝟐) 𝒅= 𝒊𝟏+ 𝒊𝟐− 𝒓𝟏−𝒓𝟐 𝒅= ( 𝒊𝟏+ 𝒊𝟐) −( 𝒓𝟏+𝒓𝟐) 𝐵𝑢𝑡 𝑨= 𝒓𝟏+𝒓𝟐 N.B: For calculations involving prisms, it is easier to consider refractions at each side separately using the law of reversibility of light. 𝒊𝟏 𝑨 𝑨 𝒅= (𝒊𝟏+ 𝒊𝟐) −𝑨 𝑤ℎ𝑒𝑟𝑒 𝑨 𝒊𝒔 𝒕𝒉𝒆 𝒓𝒆𝒇𝒓𝒂𝒄𝒕𝒊𝒏𝒈 𝒂𝒏𝒈𝒍𝒆 𝒐𝒇 𝒑𝒓𝒊𝒔𝒎 NOTE: 𝒅𝟏 𝒅𝟐 𝒓𝟏 𝒓𝟐 𝑨 𝒓𝟏 𝒄 1 ssekwerobert@gmail.com 0752053997 𝒊𝟏 If a ray of light does not emerge out but instead grazes on one surface of the prism, then the angle of refraction at that surface is equal to the critical angle. 𝒊𝟐 𝒊𝟐 𝟓𝟎𝟎 i) angle of refraction at both surfaces ii) refracting angle of the prism. iii) total deviation of the prism. Examples: 1. A ray of light is incident on a glass prism of refractive index 1.5 at an angle of 400. Given that the ray emerges out an angle of 58.40. Find the; 𝑨 𝟓𝟎𝟎 𝐬𝐢𝐧𝒓𝟏= 𝟎. 𝟒𝟐𝟖𝟓 𝒓𝟏= 𝒔𝒊𝒏−𝟏(𝟎. 𝟒𝟐𝟖𝟓) 𝒓𝟏= 𝟐𝟓. 𝟒𝟎 ii) 𝐬𝐢𝐧𝒓𝟐= 𝟎. 𝟖𝟓𝟏𝟕 𝟑𝟎𝟎 𝟓𝟖. 𝟒𝟎 SSEKWE ROBERT i) 𝐑𝐞𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐚𝐭 𝐟𝐢𝐫𝐬𝐭 𝐬𝐮𝐫𝐟𝐚𝐜𝐞 ∩𝟏𝐬𝐢𝐧𝒊= ∩𝟐𝐬𝐢𝐧𝒓