A concave mirror has radius of curvature of 40cm. It is at the bottom of a glass that has water filled up to 5cm (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 close to : (Refractive index of water =1.33) 
The correct option is A 8.8cm
Light incident from particle P will be reflected at mirror
u=−5cm,f=−R2=−20cm
1v+1u=1f
v1=+203cm
This image will act as object for light getting refracted at water surface
So, object distance d=5+203=353cm
below water surface,
After refraction, final image is at
d′=d(μ2μ1)
=(353)(14/3)
=354=8.75cm
≈8.8cm
A concave mirror has a 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 close to :
(Refractive index of water=1.33)
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 in cm from the surface of water. The value of d is close to
[Refractive index of water = 4/3]
The convex surface of a thin concavo-convex lens of the glass of refractive index 1.5 has a radius of curvature 20cm. The concave surface has a radius of curvature 60cm. The convex side is silvered and placed on a horizontal surface.
(a) Where should a pin be placed on the optic axis such that its image is formed at the same place?
(b) If the concave part is filled with water of refractive index 4/3, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.