Question

A lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = 112.4 cm. The tip of the object arrow is located at(x,y) = (x1, y1) = (-45.9 cm, 4.91 cm). The index of refraction of the lens is n = 1.43.

1) What is flens, the focal length of the lens? If the lens is converging flens is positive. It the lens is diverging, flens is negative.


2) What is y2, the y co-ordinate of the image of the tip of the arrow?

3) The lens is a plano-convex lens. What is Rlens, the radius of curvature of the convex side of the lens?

4) The object arrow is now moved to x = x1,new = -15 cm. What is x2,new, the new x co-ordinate of the image of the arrow?

5) Is the new image of the arrow real or virtual? Is it upright or inverted?

Real and upright
Real and inverted
Virtual and upright
Virtual and inverted

Answer 1

Given
coordinates of the tip of the arrow (-45.9 cm, 4.91 cm)
The object distance is u = 45.9 cm
The height of the object is h = 4.91 cm
The image distance v = x2 = 112.4 cm
The index of refraction of the material is n = 1.43

1)
The thin lens equation is
1/f = 1/u + 1/v
1/f = 1/45.9 + 1/112.4
f = 32.6 cm

The focal length of the lens is +32.6 cm

2)
The magnification M = -v/u
                                   = -112.4 cm/45.9 cm
                                   = -2.45
The image is real and inverted
The height of the image is h' = Mh
                                                = -2.45*4.91 cm
                                                = -12.03 cm

The coordinate of the tip of the image is (112.4 cm, -12.03 cm)
The Y coordinate is -12.03 cm

3)
The lens maker's formula is
1/f = (n-1)[1/R₁ - 1/R₂]
For plano-convex lens, R₂ = ∞

Therefore
1/f = (n-1)[1/R₁]
1/32.6 = (1.43-1)(1/R₁)
R₁ = 14.02 cm
The radius of curvature is 14.02 cm

4)
New coordinate of x1 = u = 15 cm
The thine lens equation is
1/f = 1/u + 1/v
1/32.6 = 1/15 + 1/v
v = -27.8 cm

the new X coordinate of the image is x2 = -27.8 cm

5)
The image is virtual and upright. (v is negative)