The crystalline lens of the human eye is a double convex lens made of material haveing an index of refraction of 1.44. Its focal length in air is about 8 mm which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) if an object 16 cm tall is placed 30 cm from the eye lens where would the lens focus it and how tall would the image be? is this image real or virtual? erect or inverted?

I cant figure out part a, I have an equation but I am confused. Help please!

The focal length f of a biconvex lens is given by

1/f = (n-1) (1/R1 - 1/R2)= 2(n-1)/R if the two radii of curvature both equal R and the curvatures are in opposite directions.

1/8 = 2*(0.44)/R

R = 7.04 cm

7.04 mm*

4.80 11 years later

To find the radii of curvature of the double convex lens, we can use the lens formula:

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

where f is the focal length of the lens, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface of the lens, and R2 is the radius of curvature of the second surface.

Given that the focal length (f) is 8 mm and the refractive index (n) is 1.44, we need to find the values of R1 and R2.

Since the problem states that the radii of curvature of its two surfaces have the same magnitude, we can assume R1 = -R and R2 = R, where R is the magnitude of the radius of curvature.

Substituting these assumptions into the lens formula, we get:

1/8 = (1.44 - 1) * ((1/(-R)) - (1/R)).

Simplifying this equation, we can solve for R:

1/8 = (0.44/R).

Cross multiplying, we get:

R = 8/0.44 = 18.18 mm.

Therefore, the radius of curvature (magnitude) for each surface of the lens is approximately 18.18 mm.

For part (b) of the question, we can use the lens formula again to determine where the lens will focus the object and the characteristics of the resulting image.

Given that the object height (h) is 16 cm and the object distance (u) is -30 cm (since the object is placed on the opposite side of the lens from the incident light), we need to find the image distance (v) and the image height (h').

Using the lens formula:

1/f = (1/v) - (1/u),

we can substitute the given values:

1/8 = (1/v) - (1/(-30)).

Simplifying this equation, we can solve for v:

1/8 = (-1/30) - (1/v).

Combining the terms, we get:

1/v = 1/8 + 1/30 = (30 + 8)/240 = 38/240.

Inverting both sides of the equation, we have:

v = 240/38 = 6.32 cm.

Therefore, the image distance is approximately 6.32 cm.

Now, to find the image height (h'), we can use the magnification formula:

magnification (m) = height of image (h') / height of object (h) = -v/u.

Substituting the given values, we get:

m = h' / h = -v/u = (-6.32) / (-30) = 0.21.

Rearranging this equation, we can solve for the image height:

h' = m * h = 0.21 * 16 cm = 3.36 cm.

Therefore, the image height is approximately 3.36 cm.

Now, let's analyze the characteristics of the image:

- The image distance (v) is positive, indicating that the image is formed on the same side of the lens as the incident light. Therefore, the image is a real image.
- The magnification (m) is positive, indicating that the image is erect.
- Since the magnification (m) is less than 1, the image height (h') is smaller than the object height (h), indicating that the image is reduced in size.

In conclusion, the lens will focus the object at a distance of approximately 6.32 cm from the lens, and the image height will be approximately 3.36 cm. The image formed is a real image, erect, and reduced in size compared to the object.