Q2. A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm. An object is situated in the denser medium at a distance of 9 cm from the pole. Locate the image due to refraction in air.
(1)  A real image at 8 cm                          (2) A virtual image at 8 cm
(3) A real image at 4.8 cm                        (4)  A virtual image at 4.8 cm

Q3.  In a medium of refractive index 1.6 and having a convex surface has a point object in it at a distance of 12 cm from the pole. The radius of curvature is 6 cm. Locate the image as seen from air.
(1)  A real image at 30 cm                          (2) A virtual image at 30 cm
(3) A real image at 4.28 cm                        (4)  A virtual image at 4.28 cm
 

Dear Student ,
Here in this case the refraction at a concave surface is given by ,
n2v-n1u=n2-n1RWhere n1=refracting index of the refracting medium=1·5 ,n2=refracting index of the medium in which the light was travelling initiallyusually air=1,v=image distance ,u=object distance=9 cm ,R=radius of curvature of the refracting surface=12 cm .Now substituting this values in the above equation we get ,1v-1·59=1-1·5121v=1·59-1-1·512v=8 cmNow the image will be virtual image and the image distance will be 8 cm .
Hope this helps you .
Please ask one question at a time so the experts can answer you properly .
Regards

  • -73
What are you looking for?