Question

what are the radii of a symmetric converging plastic lens (i.e., two equally curved surfaces) that will form an image on the screen twice the height of the object? (take the index of refraction of plastic to be 1.59.)

Answer 1

To find the radii of a symmetric converging plastic lens that will form an image on the screen twice the height of the object, we can use the Lens Maker's Equation, lens formula, and magnification formula. We can assign the radii of curvature of the lens surfaces as equal and use the given conditions to solve for the focal length and radii of curvature.

Step-by-step explanation:

To find the radii of a symmetric converging plastic lens that will form an image on the screen twice the height of the object, we can use the Lens Maker's Equation. The equation states that 1/f = (n2 - n1) * (1/R1 - 1/R2), where f is the focal length, n2 is the refractive index of the lens, n1 is the refractive index of the surrounding medium (in this case, air with a refractive index of 1), R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.

Since the lens is symmetric with two equally curved surfaces, we can take the radii of curvature to be the same, so we assign the same value to R1 and R2. Let's call this value R. We are given that the height of the image is twice the height of the object, so the magnification is given by -h2/h1 = -2.

Using the magnification formula, we have -h2/h1 = d2/d1 = -2. Given the object distance do and image distance d1, we can use the lens formula 1/f = 1/do - 1/d1, where f is the focal length of the lens, do is the object distance, and d1 is the image distance. Since d2 = -2d1, we substitute d2 = -2d1 in the lens formula to solve for f. Finally, we can use the Lens Maker's Equation to find the radius of curvature R.

Answer 2

The radius of curvature of the lens is
`\(7.08 \, \text{cm}\)`

How to determine this?

Given:

Magnification
`(\(m\)) = 2 (\(d_i/d_o\))`

Object distance
`(\(d_o\)) = 27.0 cm`

Using magnification formula:

`\(d_i = 2 * d_o = 2 * 27.0 \, \text{cm} = 54.0 \, \text{cm}\)`

Therefore, the image distance
`(\(d_i\))` is 54.0 cm, and the object distance
`(\(d_o\))` is 27.0 cm.

Now, finding the focal length (\(f\)) using the lens formula:

`\((1)/(f) = (1)/(d_o) + (1)/(d_i)\)`

`\((1)/(f) = \frac{1}{27.0 \, \text{cm}} + \frac{1}{54.0 \, \text{cm}}\)`

`\((1)/(f) = \frac{1}{9.0 \, \text{cm}} + \frac{1}{18.0 \, \text{cm}}\)`

`\((1)/(f) = \frac{3}{18.0 \, \text{cm}}\)`

`\(f = 6.0 \, \text{cm}\)`

Now, applying the lens equation:

`\((1)/(f) = (n - 1) \cdot \left((1)/(R_1) + (1)/(R_2)\right)\)`

Given that
`\(R_1 = R_2 = R\)`,

`\((1)/(f) = (n - 1) \cdot \left((2)/(R)\right)\)`

Given
`\(f = 6.0 \, \text{cm}\), \(n = 1.59\)`:

`\(\frac{1}{6.0 \, \text{cm}} = (1.59 - 1) \cdot \left((2)/(R)\right)\)`

`\(\frac{1}{6.0 \, \text{cm}} = 0.59 \cdot \left((2)/(R)\right)\)`

`\(R = \frac{0.59 \cdot 2 \cdot 6.0 \, \text{cm}}{1} = 7.08 \, \text{cm}\)`

Therefore, the radius of curvature of the lens is
`\(7.08 \, \text{cm}\)`.

Complete question:

An object is 27.0 cm from a screen.

What are the radii of a symmetric converging plastic lens (i.e., two equally curved surfaces) that will form an image on the screen twice the height of the object? (Take the index of refraction of plastic to be 1.59.)