Saved Bookmarks
| 1. |
(a) Fig. 9.04 shows a cross-section of a Tight pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.(b) What is the answer if there is no outer covering of the pipe ? |
|
Answer» Solution : (a) Here refractive index of glass fibre `n_g = 1.68` and refractive index of OUTER covering `n_c = 1.44` Total internal reflection will take place inside the pipe if angle of incidence `i. gt i_( c)` , the critical angle, for glass fibre-outer covering interface. But `sin i_( c) =1/(n_(ge)) =n_( c)/n_(G) = 1.44/1.68 = 0.855 rArr i_( c) sin^(-1) (0.855) = 59^(@)` Hence, for total reflection `i^(.) gt 59^(@)`, or R, which has a value (90 -i.), should have a value `r le (90 -59)` or `r le 31^(@)`. Now according to Snell.s law `(sin i)/(sin r) = n_(g)` `therefore sin i = n_(g). sin r =(1.68) xx sin 31^(@) = 1.68 xx 0.5150 = 0.8652` `therefore i = sin^(-1) (8652) = 60^(@)` Hence, range of the ANGLES of incident rays with the axis of pipe = less than or equal to ± 60°. If there is no outer covering, then air will act as the covering and hence the critical angle `i_( c)^(.) = sin^(-1) (1/n_(g)) = sin^(-1) (1/1.68) = 36.5^(@)` Now, for rays incident along the axis of pipe i = 90° and applying Snell.s law `(sin i)/(sin r) = (sin 90^(@))/(sin r) = 1/(sin r) = n_(g) = 1.68`, we find that r= `36.5^(@)` Consequently, `i. = 90-r - 36.5 = 53.5^(@)`and it has a value greater than the critical angle i.c. Hence, all the rays will be totally reflected. |
|