Question
An equi-convex lens of refractive index μg = 1.5 and focal length 10cm is placed on the surface of water (μw = 4/3) such that its lower surface is immersed in water but its upper surface is in contact with air outside, (a) At what distance from the lens will a beam parallel to its principal axis come to focus? (b) How is the position of the focus altered if the lens is wholly immersed in water?
Solution
Let, the radius of curvature of each face of the lens be x cm. Then R1 = + x cm and R2= – x cm. Also,
Refractive index f glass, μg = 1.5
Refractive index of water, = 4/3
Focal length, f = + 10 cm.
Substituting inthe lens makers formula,
We have,
which gives,
x = 10 cm
Thus, R1 = + 10 cm and R2 = –10 cm.
(a) As shown in the figure above, I is the image of object O due to refraction at the upper face.
Since the refractive index of the incident ray which is in air is given by μa and the refracted ray in glass is of refractive index , we have
refractive index on this surface (whose radius of curvature is given by R1)
...(i)
where,
The image I’ serves as the virtual object for refraction at the lower surface.
For refraction at this surface, the incident ray is in glass and the refracted ray in water and I is the final image.
Thus, for refraction at the lower surface (whose radius of curvature is R2) we have
...(ii)
Adding (i) and (ii) we get
Now, substituting the values,
(b) If the lens is wholly immersed in water, the formula is
which (for u = – ∞) gives v = 40 cm.
The focus of the rays when immersed in water increases by 20 cm.
Refractive index f glass, μg = 1.5
Refractive index of water, = 4/3
Focal length, f = + 10 cm.
Substituting inthe lens makers formula,
We have,
which gives,
x = 10 cm
Thus, R1 = + 10 cm and R2 = –10 cm.
(a) As shown in the figure above, I is the image of object O due to refraction at the upper face.
Since the refractive index of the incident ray which is in air is given by μa and the refracted ray in glass is of refractive index , we have
refractive index on this surface (whose radius of curvature is given by R1)
...(i)
where,
The image I’ serves as the virtual object for refraction at the lower surface.
For refraction at this surface, the incident ray is in glass and the refracted ray in water and I is the final image.
Thus, for refraction at the lower surface (whose radius of curvature is R2) we have
...(ii)
Adding (i) and (ii) we get
Now, substituting the values,
(b) If the lens is wholly immersed in water, the formula is
which (for u = – ∞) gives v = 40 cm.
The focus of the rays when immersed in water increases by 20 cm.
