A biconcave lens made from glass with n=1.5. The radii of curvatures are

20cm and 10cm, and thickness is 5cm. An object of height 2cm is placed 8cm to the left of surface of the lens.

a) Describe the final image; location, nature (real/virtual), size, orientation (erect/inverted).

b) Answer part (a) by using the thin lens approximation. Compare your results with the results found previously and comment.

To answer both parts of the question, we can use the lens formula and magnification formula. The lens formula relates the object distance (u), focal length (f), and image distance (v) as follows:

1/f = (n - 1) * (1/R1 - 1/R2),

where n is the refractive index of the lens material, R1 is the radius of curvature of the first surface (positive for convex, negative for concave), and R2 is the radius of curvature of the second surface.

The magnification formula is given by:

magnification (m) = -v/u,

where negative magnification indicates an inverted image.

a) Describing the final image:
1. Calculate the focal length (f) using the lens formula.
2. Calculate the image distance (v) using the lens formula with the given object distance (u = -8cm).
3. Calculate the magnification (m) using the magnification formula.
4. Determine the nature (real/virtual) of the image based on the image distance.
5. Determine the size and orientation of the image based on the magnification.

b) Answering part (a) using the thin lens approximation:
1. Determine the effective focal length (f') using the thin lens approximation.

Now let's go step by step through each part:

a) Describing the final image:

1. Calculate the focal length (f) using the lens formula:
Given that n = 1.5, R1 = 20cm, and R2 = 10cm, we substitute these values into the lens formula:
1/f = (1.5 - 1) * (1/20 - 1/10)
1/f = 0.5 * (-1/20)
1/f = -1/40
f = -40cm (focal length is negative for a biconcave lens)

2. Calculate the image distance (v) using the lens formula with the given object distance (u = -8cm):
1/f = (1.5 - 1) * (1/R1 - 1/R2)
1/-40 = 0.5 * (1/20 - 1/10)
-1/40 = 0.5 * (-1/20)
v = (-40cm * -8cm) / (-40cm - -8cm)
v = 320cm / 32cm
v = 10cm

3. Calculate the magnification (m) using the magnification formula:
m = -v/u
m = -10cm / -8cm
m = 1.25

4. Determine the nature (real/virtual) of the image based on the image distance:
Since the image distance (v) is positive (+10cm), the image is virtual.

5. Determine the size and orientation of the image based on the magnification:
The magnification (m) is positive (1.25), indicating an erect image. The size of the image can be determined by multiplying the magnification by the object height:
Image height = magnification * object height
Image height = 1.25 * 2cm
Image height = 2.5cm

Therefore, the final image is virtual, erect, and has a height of 2.5cm.

b) Answering part (a) using the thin lens approximation:

1. Determine the effective focal length (f') using the thin lens approximation:
The thin lens approximation assumes that the thickness of the lens is negligible compared to the radii of curvature. In this case, the lens is biconcave, so the effective focal length can be approximated as:
1/f' = (n - 1) * (1/R1 - 1/R2)
1/f' = (1.5 - 1) * (1/20 - 1/10)
1/f' = 0.5 * (-1/20)
1/f' = -1/40
f' = -40cm

Comparing the result obtained from the thin lens approximation (f' = -40cm) with the previous result using the lens formula (f = -40cm), we can see that they are the same. This indicates that the thin lens approximation is a valid and accurate way to determine the focal length for this biconcave lens.

In conclusion, both methods yield the same results for the final image. The image is virtual, erect, and has a height of 2.5cm.