Question

Part (a) of the figure below shows the eye of a person looking through a lens so that the index of refraction of the medium that the incident light is moving through is ni 1.2 and the person's eye is in a medium having an index of refraction given by 1.8, which is also the index of refraction of the medium that the refracted light moves through into the eye of the person. of the lens if the person moves their eye to the other side of the lens, as shown in part (b) of the figure. Note that here, the index of refraction of the medium that the incident light is moving through is now n1.8 and the person's eye is in a medium having an index of refraction given by nR 1.2, which is also the index of refraction of the medium that the refracted light moves through into the eye of the person

Answer 1

n/v - 1.2/ infinity = n - 1.2/R_1 1.8/f_a - n/v = 1.8 - n/ - R_2 rightarrow (2) (1) rightarrow (2) 1.8/f_a - 1.2/infinity = n - 1.2/R_1 + n - 1.8/R_2 [If R_1 = R_2] 1.8/360 = 2n - 3.0/R (I) n/v - 1.8/infinity = n - 1.8/R 1.2/f_b - n/v = n - 1.2/R/1.2/f_b = 2n - 3.0/R = 1.8/360 from eq(I) Therefore f_b = 360/1.8 times 1.2 = 240cm