35) A convex lens forms a real image of an object on a screen; the magnification of the image being 3/2. The object and the screen are kept fixed and the lens is moved through a distance of 16 cm when a sharp image is again formed on the screen; the magnification now being 2/3. What is the focal length of the lens?
19.2 cm
Related Optics MCQ with Answers
1.55
n1/u +n2/v=(n2-n1)/r for first refraction by convex surface: u??, v=rn2/(n2-n1) for plane reflection: r is infinite,v'=-v for second refraction: 1.5/v' -1/v''=(1-1.5)/r 1/v''+1/v=1/f power=1/f
11 cm
The curved silvered surface will behave as a concave lens of focal length fm=R/2=?22/2 =?11cm =?0.11m And hence PM=the power of mirror =?1/fM=?1/?0.11=1/0.11D Further as focal length of lens is 20 cm ie 0.2m its power is P2=1/f2P2 =1/20D Now as in image formation, light after passing through the lens will be reflected back by curved mirror through lens again . P=PL+PM+PL =2PL+PM P=2/0.2+1/0.11=210/11D So, focal length of equivalent mirror. F=?1/P=11/210m ?110/21cm ie silvered lens behaves as a concave mirror of focal length (110/21)cm So far object at a distance 10 cm in front of it. 1/V+1/?10=21/110 i,e V=?11cm The image will be 11 cm in front of silvered lens and will be real. Hence c is the correct answer.
2 cm and 12 cm
The most commonly used formula in amateur astronomy is used to calculate the magnification of a telescope: magnification = focal length of telescope / focal length of eyepiece. Example: using a 10mm eyepiece in a telescope with a focal length of 1000mm results in a magnification of 100x (1000 / 10 = 100)