1) A convex lens with a focal lenght of 4.0cm is placed 20.0cm from a concave lens with a focal lenght of 5.0cm. Find the position of the image when the object is placed 12.0c, in front of the convex lens.

2) A double concave lens has a refractive index of 1.5 radii of curvature of 10.0cm and 15cm. Determine its focal length.

Thank you for helping in advance!

1) Use the position of the image formed by the first lens as the location of the object for the second lens.

For the distance of the image from the first lens, di, use
1/12 + 1/di = 1/4
1/di = 2/12
di = 6
That means the first image is 20 - 6 = 14 cm in front of the concave lens, which has a negative focal length. For the location of the final image, di', use
1/14 + 1/di' = -1/5

2) For this problem, use the "lens maker's formula". If it is not in your textbook, you can find it at

http://scienceworld.wolfram.com/physics/LensMakersFormula.html

Both of the "1/R" terms will be negative. The focal length will be negative.

Thank you drwls for your help!

Question 1:

To find the position of the image formed when the object is placed in front of the convex lens, we can use the lens formula:

1/f = 1/v - 1/u

Where:
f = focal length of the lens
v = distance of the image from the lens
u = distance of the object from the lens

Given:
fconvex = 4.0 cm (focal length of the convex lens)
fconcave = 5.0 cm (focal length of the concave lens)
u = -12.0 cm (object distance from the convex lens)

First, we need to find the distance between the two lenses. Since the concave lens is located 20.0 cm from the convex lens, the object distance for the concave lens (uconcave) is equal to 20.0 cm - u.

uconcave = 20.0 cm - (-12.0 cm)
= 20.0 cm + 12.0 cm
= 32.0 cm

Now we can find the image distance (vconcave) using the lens formula for the concave lens:

1/fconcave = 1/vconcave - 1/uconcave

Rearranging the formula, we get:

1/vconcave = 1/fconcave + 1/uconcave

Plugging in the values, we have:

1/vconcave = 1/5.0 cm + 1/32.0 cm

Now, we can solve for vconcave:

1/vconcave = (32.0 cm + 5.0 cm) / (5.0 cm * 32.0 cm)
= 37.0 cm / 160.0 cm
= 0.23125 cm^(-1)

Taking the reciprocal of both sides, we find:

vconcave = 1 / (0.23125 cm^(-1))
= 4.33 cm

Since the position of the image for the convex lens is the same as the position of the object for the concave lens, the final image position is v = 4.33 cm.

Therefore, the position of the image when the object is placed 12.0 cm in front of the convex lens is 4.33 cm.

Question 2:

To determine the focal length of the double concave lens, we can use the lensmaker's formula:

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

Where:
f = focal length of the lens
n = refractive index of the lens material
R1 = radius of curvature of the first surface
R2 = radius of curvature of the second surface

Given:
n = 1.5 (refractive index of the lens)
R1 = 10.0 cm
R2 = 15.0 cm

Plugging in the values, we have:

1/f = (1.5 - 1) * (1/10.0 cm - 1/15.0 cm)

Performing the calculations:

1/f = (0.5) * (0.15 cm^(-1) - 0.10 cm^(-1))
= (0.5) * (0.05 cm^(-1))
= 0.025 cm^(-1)

Taking the reciprocal of both sides, we find:

f = 1 / (0.025 cm^(-1))
= 40.0 cm

Therefore, the focal length of the double concave lens is 40.0 cm.