One end of a cylindrical glass rod of refractive index 1.5 is a hemispherical surface of radius of curvature 20 mm. An object is placed on the axis of the rod at 80 mm to the left of the vertex of the surface. Determine the position if the image of the rod is immersed in water of refractive index 1.33
Question 3 Answer
To find the position of the image, we can use the lens formula:
1/f = (n2 - n1) * (1/r1 - 1/r2)
where:
f is the focal length of the lens (in this case, the hemispherical surface),
n1 is the refractive index of the surrounding medium (air),
n2 is the refractive index of the lens (glass),
r1 is the radius of curvature of the initial surface (positive for convex surface),
and r2 is the radius of curvature of the final surface (positive for convex surface).
Given:
n1 = 1, the refractive index of air,
n2 = 1.5, the refractive index of the glass,
r1 = -20 mm (negative because it is a convex surface),
r2 = +∞ for the final surface (since it is in contact with water).
Substituting these values into the lens formula, we have:
1/f = (1.33 - 1) * (1/-20 - 1/∞)
Since 1/∞ is zero, the equation simplifies to:
1/f = 0.33 * (-1/20) = -0.0165
Solving for f, we get:
f = -1/0.0165 ≈ -60.61 mm
The negative sign indicates that the lens is diverging.
Now, using the magnification formula:
m = -d1/d2
where:
m is the magnification,
d1 is the object distance (80 mm),
and d2 is the image distance.
Rearranging the formula, we have:
d2 = -d1/m
Substituting the values, we get:
d2 = -80 / -60.61 ≈ 1.32 mm
Therefore, the position of the image of the object is approximately 1.32 mm to the right of the hemispherical surface.