a planoconvex lens has a max thickness of 6cm.when placed on q horizon...
Explanation of the problem
The problem involves a planoconvex lens, which means that one side of the lens is flat (plane) while the other side is curved (convex). The lens has a maximum thickness of 6cm, and when it is placed on a horizontal table with the curved surface in contact with the surface, the bottommost point of the lens appears to be at a depth of 4cm. When the lens is inverted such that the plane face is in contact with the table surface, the center of the plane face appears to be at a depth of 17/3cm.
Solution
To solve this problem, we can use the concept of refraction of light. When light passes through a lens, it changes direction due to the change in the refractive index of the medium. The amount of bending of light depends on the curvature of the lens surface and the refractive index of the lens material.
Step 1: Find the refractive index of the lens
We are not given the refractive index of the lens material, so we need to find it. We can do this by using Snell's law of refraction, which states that:
n1 * sinθ1 = n2 * sinθ2
where n1 and n2 are the refractive indices of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.
In this case, we can assume that the lens is made of glass, which has a refractive index of around 1.5. We also know that the angle of incidence is 0 degrees, since the lens is in contact with the table surface. Therefore, we can simplify Snell's law to:
n1 * sin(0) = n2 * sinθ2
n1 = n2 * sinθ2
Since the angle of refraction depends on the curvature of the lens surface, we need to find the radius of curvature of the lens next.
Step 2: Find the radius of curvature of the lens
We can use the apparent depth of the bottommost point of the lens to find the radius of curvature. When the curved surface of the lens is in contact with the table, the bottommost point of the lens appears to be at a depth of 4cm. This means that the light rays from that point appear to be coming from a virtual image located 4cm below the table surface.
Using the lens formula, we can relate the distance of the virtual image (vi) to the distance of the object (vo) and the focal length (f) of the lens:
1/f = 1/vi + 1/vo
Since the object is at a distance of 6cm (the thickness of the lens), we can simplify the lens formula to:
1/f = 1/vi + 1/6
We know that the virtual image appears to be at a depth of 4cm, so vi = -4. Substituting this value into the lens formula gives:
1/f = -1/4 + 1/6
f = -12
This means that the lens has a focal length of -12cm, which implies that it is a diverging lens (since the focal length is negative for a diverging lens).
The radius of curvature