Question

A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?

Short answer

Answer: The photographer should use a lens with a focal length of approximately 19.96 mm.

Step-by-step solution

Write down the given information

The height of the Eiffel Tower is 300 m, the distance from the photographer to the Tower (object distance) is 300 m, and we need an image 20 mm high on the film. We must find the focal length of the lens.

Determine the object distance (d_o) and image distance (d_i)

The object distance (d_o) is the distance from the lens to the object, which is 300 m.

Determine the magnification formula

The magnification formula is: magnification = (image height) / (object height) = (-d_i) / (d_o)

Solve for the image height

The image height is given as 20 mm. Since we're dealing with distances in meters, let's convert the image height to meters: 20 mm = 0.02 m.

Calculate the magnification

Now we will calculate the magnification using the formula. magnification = (0.02 m) / (300 m) = 1 / 15000

Determine the thin lens formula

The thin lens formula is: 1 / f = 1 / d_o + 1 / d_i

Use magnification and thin lens formulas to solve for f

Since magnification = (-d_i) / (d_o), we can rewrite this equation in terms of d_i: d_i = -d_o * magnification Now plug in the values for d_o and magnification: d_i = -300 m * (1 / 15000) = -0.02 m The image will be 0.02 m behind the lens. Now, use the thin lens formula to find the focal length: 1 / f = (1 / 300) + (1 / (-0.02))

Solve for the focal length

We will now solve for the focal length, f: 1 / f = (1 / 300) - (1 / 0.02) f = 1 / ((1 / 300) - (1 / 0.02)) f ≈ 0.01996 m or 19.96 mm The photographer should use a lens with a focal length of approximately 19.96 mm to get an image of the Eiffel Tower that is 20 mm high on the film.