A thin converging lens of focal length 9.82 cm

forms an image of an object placed 18.5 cm
from the lens.
Find the image distance.
Answer in units of cm.

What is the magnification for an object distance of 18.5 cm?

Find the location of the image for an object
distance of 4.06 cm.
Answer in units of cm.

Calculate the magnification for an object distance of 4.06 cm.

You missed a very important lecture. The lens law.

Do = object distance
Di = image dstance
f = focal length

1/Do + 1/Di = 1/f

magnification = |Di/Do|

Try using those formulas for these questions. You need to learn how to use them to pass your course.

The first answer is 20.93, then when I take that and divide it by 18.5 equals 1.13 but that is incorrect. What am I doing wrong? Then using the same formula (1/9.82)- (1/4.06) comes out -6.92 which doesn't seem right either. please help.

I was able to get distance but both my magnifications were incorrect. pls help

Got it ~ thanks!

To find the image distance for a thin converging lens, we can use the lens formula:

1/f = 1/v - 1/u

Where:
f is the focal length of the lens
v is the image distance
u is the object distance

For the first question, we are given:
f = 9.82 cm
u = 18.5 cm

Plugging these values into the lens formula, we can solve for v:

1/9.82 = 1/v - 1/18.5

Rearranging the equation, we get:

1/v = 1/9.82 + 1/18.5

Adding the fractions on the right side:

1/v = (18.5 + 9.82) / (9.82 * 18.5)

Simplifying:

1/v = 28.32 / 181.57

Now, taking the reciprocal of both sides:

v = 181.57 / 28.32

Calculating this expression, we find:

v ≈ 6.41 cm

Therefore, the image distance for an object distance of 18.5 cm is approximately 6.41 cm.

To calculate the magnification for this object distance, we can use the formula:

Magnification (m) = -v/u

Plugging in the values:

m = -6.41 / 18.5

Calculating this expression, we find:

m ≈ -0.35

The magnification for an object distance of 18.5 cm is approximately -0.35.

For the second question, we are given:
u = 4.06 cm

Using the lens formula again, we can find the image distance:

1/9.82 = 1/v - 1/4.06

Rearranging the equation:

1/v = 1/9.82 + 1/4.06

Adding the fractions:

1/v = (4.06 + 9.82) / (9.82 * 4.06)

Simplifying:

1/v = 13.88 / 39.97

Taking the reciprocal:

v = 39.97 / 13.88

Calculating this expression, we find:

v ≈ 2.88 cm

Therefore, the image distance for an object distance of 4.06 cm is approximately 2.88 cm.

To calculate the magnification for this object distance, we use the same formula:

m = -v/u

Plugging in the values:

m = -2.88 / 4.06

Calculating this expression, we find:

m ≈ -0.71

The magnification for an object distance of 4.06 cm is approximately -0.71.