Question

A person on a safari wants to take a photograph of a hippopotamus from a distance of \(75.0 \mathrm{m} .\) The animal is \(4.00 \mathrm{m}\) long and its image is to be \(1.20 \mathrm{cm}\) long on the film. (a) What focal length lens should be used? (b) What would be the size of the image if a lens of \(50.0-\mathrm{mm}\) focal length were used? (c) How close to the hippo would the person have to be to capture a \(1.20-\mathrm{cm}-\) long image using a 50.0 -mm lens?

Short answer

Answer: The person needs a 76.3 mm focal length lens to capture the desired image size. If a 50.0-mm lens is used, the person would have to be 50.8 cm away from the hippo to capture a 1.20 cm long image.

Step-by-step solution

(a) Finding the focal length

First, we need to find the image distance (\(d_i\)). We can use the magnification formula and the relationship between magnification, object size, and image size: $$M = h_i / h_o$$ Substituting the given values: $$M = \frac{1.20 \thinspace cm}{4.00 \thinspace m} = \frac{1.20 \thinspace cm}{400 \thinspace cm} = 0.003$$ Now we can use the magnification formula to find the image distance (\(d_i\)): $$M = -(d_i / d_o)$$ $$d_i = -Md_o = -(0.003)(75.0 \thinspace m) = -0.225 \thinspace m$$ Using the thin lens formula, we can find the focal length (\(f\)): $$(1/f) = (1/d_o) + (1/d_{i})$$ Substituting the values and solve for \(f\): $$\frac{1}{f} = \frac{1}{75.0 \thinspace m} + \frac{1}{-0.225 \thinspace m}$$ $$\frac{1}{f} = 0.0131 \thinspace m^{-1}$$ $$f = 1/0.0131 = 76.3 \thinspace mm$$ Therefore, the person needs a \(76.3 \thinspace mm\) focal length lens.

(b) Finding the image size with a 50.0-mm lens

Now let's find the size of the image if a lens of \(50.0 \thinspace mm\) focal length is used. First, we find the image distance (\(d_i\)) using the thin lens formula: $$(1/f) = (1/d_o) + (1/d_{i})$$ Plug in the given values and solve for \(d_i\): $$\frac{1}{50.0 \thinspace mm} = \frac{1}{7500 \thinspace mm} + \frac{1}{d_i}$$ $$\frac{1}{d_i} = \frac{1}{50.0 \thinspace mm} - \frac{1}{7500 \thinspace mm}$$ $$d_i = 53.6 \thinspace mm$$ Now we can find the magnification \(M\) for the new lens using the magnification formula: $$M = -(d_i / d_o)$$ $$M = - \frac{53.6 \thinspace mm}{7500 \thinspace mm} = -0.00715$$ Now, we can find the image size by using the relationship between magnification, object size, and image size: $$h_i = Mh_o$$ $$h_i = (-0.00715)(4 \thinspace m) = -0.286 \thinspace cm$$ Therefore, the image size would be \(0.286 \thinspace cm\) (negative sign indicates an inverted image) if a lens of \(50.0 \thinspace mm\) focal length were used.

(c) Finding the required object distance to obtain the desired image using a 50.0-mm lens

Now we need to find how close the person would have to be to the hippo to capture a \(1.20 \thinspace cm\) long image using a \(50.0 \thinspace mm\) lens. We can use the magnification formula and the relationship between object size, image size, and magnification that we used in part (a). First, find the required magnification: $$M = h_i / h_o = \frac{1.20 \thinspace cm}{4.00 \thinspace m} = 0.003$$ We now use the magnification formula to find the required object distance (\(d_o\)): $$M = -(d_i / d_o)$$ Next, we need to find the image distance (\(d_i\)) using the thin lens formula with the given focal length of \(50.0 \thinspace mm\): $$(1/f) = (1/d_o) + (1/d_{i})$$ $$\frac{1}{50.0 \thinspace mm} = \frac{1}{d_o} + \frac{1}{-0.003d_o}$$ Solve for \(d_o\): $$d_o = 50.8 \thinspace cm$$ So, the person would have to be \(50.8 \thinspace cm\) away from the hippo to capture a \(1.20 \thinspace cm\) long image using a \(50.0 \thinspace mm\) lens.