Question

A 2.0-cm-diameter spider is 2.6 m from a wall.

Determine the focal length f and position s' (measured from the wall) of a lens that will make a double-size image of the spider on the wall.

Answer 1

The focal length of the lens is 8.00 cm. To determine the position of the lens that will make a double-size image of the spider on the wall, we can use the lens formula and the magnification formula.

Step-by-step explanation:

The focal length of the lens is the distance from the center of the lens to the spot, given to be 8.00 cm. Thus, f = 8.00 cm. To determine the position s', we can use the lens formula:

1/f = 1/s + 1/s'

Since we want a double-size image, the magnification m = -2. We can use the magnification formula:

m = -s'/s

Substituting the known values, we can solve for s'.

Answer 2

To make a double-size image of the spider on the wall, we need a lens with a focal length of 8.00 cm and a position s' of 8.33 m from the wall.

Step-by-step explanation:

To make a double-size image of the spider on the wall, we can use the lens formula:

1/f = 1/s + 1/s',

where f is the focal length, s is the object distance, and s' is the image distance measured from the wall.

Given that the diameter of the spider is 2.0 cm and it is 2.6 m from the wall, we can calculate the object distance as s = 2 * radius = 2 * 1.0 cm = 2.0 cm (since the diameter is given). Converting it to meters, we have s = 0.02 m.

Substituting the values into the lens formula, we get:

1/8.00 = 1/0.02 + 1/s',

Simplifying the equation, we find:

1/s' = 1/8.00 - 1/0.02 = 0.12,

Therefore, the image distance measured from the wall, s', is 8.33 m.