Attempt any 4 sub parts from each question. Each question carries 1 mark.
Case Study Question 1:
A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.
Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.
(i) The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be
(a) 3.45 cm
(b) 5 cm
(c) 1.29 cm
(d) 2.59 cm
(ii) How far from the objective should an object be placed in order to obtain the condition described in part (i)?
(a) 4.5 cm
(b) 2.5 cm
(c) 1.5 cm
(d) 3.0 cm
(iii) What is the magnifying power of the microscope in case of least distinct vision?
(iv) The intermediate image formed by the objective of a compound microscope is
(a) real, inverted and magnified
(b) real, erect and magnified
(c) virtual, erect and magnified
(d) virtual, inverted and magnified
(v) The magnifying power of a compound microscope increases with
(a) the focal length of objective lens is increased and that of eye lens is decreased.
(b) the focal length of eye lens is increased and that of objective lens is decreased.
(c) focal length of both objects and eye-piece are increased.
(d) focal length of both objects and eye-piece are decreased.