Question

Please note that the glass rod is submerged in water, thus changing the focal length.

3. The left end of a long glass rod of index 1.666 is polished to a convex surface of radius 1.0 cm and then submerged in water (n = 1.333). A small object 3.0 cm high is located in the water and on the axis 12.0 cm from the vertex. Calculate (a) the primary and secondary focal lengths, (b) the power of the surface, (c) the image distance, and (d) the size of the image. [30 points]

Answer 1

2.) curvature of left end of rod= 1cm R = Icm = 0.0 m other end is plans, It means R2= مه 1.666 Refractive index of glass Refractive index of water = 1.333 Relative Refractive index of glass with respect to water = llyw lug Ugw= = 1-666 1.333 ual = 1:249 Let & is focal length of rod in water tot = (Agn-1) (1 2 1 2 こ -14 (1-249-1 0.01 f = f s 0.0 Yol m 0.249 f = 4001 cm Primary and secondary Wil -4001 cm, 4.01cm.

b.) Power of surfaces= {com) P. L 0.oyol 24.93 diopte. -1200 C.) Distance of object from rod vertex o (u) = focal les length = 4.olcm Distance of image I v By formula Ļ to = 3 V = 1 + 1 V 4.01 -12 12 ch 4.01 4.01X 12 U= 6.02 cm d) Magnification m = V 6 Where size of object = h=3cm b. и

h = size of image 1 FIR b= 3x6.02 -12 s. - - 1.505 size of image