Question

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.Part AFind the focal length of the lens that produces the image described in the problem introduction using the thin lens equation.Express your answer in centimeters, as a fraction or to three significant figures.f = 6.67 cm SubmitMy AnswersGive UpCorrectPart BConsidering the sign of f, is the lens converging or diverging?Considering the sign of , is the lens converging or diverging?convergingdivergingSubmitMy AnswersGive UpCorrectPart CWhat is the magnification m of the lens?Express your answer as a fraction or to three significant figures.m = -1.25SubmitMy AnswersGive UpCorrectPart DThink about the sign of s′ and the sign of y′, which you can find from the magnification equation, knowing that a physical object is always considered upright. Which of the following describes the nature and orientation of the image?Think about the sign of and the sign of , which you can find from the magnification equation, knowing that a physical object is always considered upright. Which of the following describes the nature and orientation of the image?real and uprightreal and invertedvirtual and uprightvirtual and invertedSubmitMy AnswersGive UpCorrectNow consider a diverging lens with focal length f=−15cm, producing an upright image that is 5/9 as tall as the object.Part EIs the image real or virtual? Think about the magnification and how it relates to the sign of s′.Is the image real or virtual? Think about the magnification and how it relates to the sign of .realvirtualSubmitMy AnswersGive UpCorrectPart FWhat is the object distance? You will need to use the magnification equation to find a relationship between sand s′. Then substitute into the thin lens equation to solve for s.Express your answer in centimeters, as a fraction or to three significant figures.s = 12.0 cm SubmitMy AnswersGive UpCorrectPart GWhat is the image distance?Express your answer in centimeters, as a fraction or to three significant figures.s′ = 24 cm SubmitMy AnswersGive UpIncorrect; Try Again; 12 attempts remaining; no points deductedA lens placed at the origin with its axis pointing along the x axis produces a real inverted image at x=−24cmthat is twice as tall as the object.Part HWhat is the image distance?Express your answer in centimeters, as a fraction or to three significant figures.s′ = 24.0 cm

Answer 1

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.