If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :

A transparent cube of side, made of a material of refractive index $$\mu $$₂, is immersed in a liquid of refractive index $$\mu $$₁($$\mu $$₁ < $$\mu $$₂). A ray is incident on the face AB at an angle $$\theta $$(shown in the figure). Total internal reflection takes place at point E on the face BC.
Then $$\theta $$ must satisfy :

A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is dose to: (Refractive index of water = 1.33)

The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the focal length of the lens used?