use m=v/v, if=1/v + 1/u
52) A spherical surface of radius of curvature R separates air (refractive index from glass (refractive index 1.5 ). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at point O and PO = OQ. The distance PO is equal to
5 R
Related Optics MCQ with Answers
-1.5 D
all power of all lens
convergent lens of focal length 3.5 R
On substituting the given quantities in the lens maker?s equation, 1/f = (n2/n1 ? 1)(1/R1 ? 1/R2), we have 1/f = [(1.5/1.75)-1] [1/(-R) ? 1/R)] = (-0.25/1.75) (-2/R) = 0.5/1.75R, from which f = 3.5R. You should ensure that all your doubts on applying the sign convention are cleared before proceeding to the next question. In the lens maker?s equation, we applied negative sign to the radius of curvature (R) of the first surface since the concave lens presents a concave surface to the incident ray (and hence the centre of curvature is on the same side of the pole as the object is). The sign of the radius of curvature of the second surface is positive since it presents a convex surface to the ray proceeding towards it after the refraction at the first surface. We did not bother about the sign of the focal length of the lens, since it is the unknown quantity in this problem. The sign turns out to be positive, indicating that the concave lens behaves as a converging lens on immersing it in a denser liquid.
Distances measured in the same direction as the incident light are taken as positive
According to Cartesian sign convention,Distances measured in the same direction as the incident light are taken as positive
hygrometer
hygrometer is used to measure humidity