A transparent photographic slide is placed in front of a converging lens with a focal length of 2.41 cm. An image of the slide is formed 13.0 cm from the slide.

(a) How far is the lens from the slide if the image is real? (Enter your answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks.)(b) How far is the lens from the slide if the image is virtual? (Enter your answers from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks.)

1/q + 1/p = 1/f

q can be +13 or -13

I couldn't figure out a, but b is (1/2.41) - (1/-13) = 2.03 cm and the second blank is none.

To solve this problem, we can use the lens formula:

1/f = 1/v - 1/u

where:
f = focal length of the lens
v = image distance
u = object distance

(a) If the image is real:
Given:
f = 2.41 cm
v = 13.0 cm

Substituting these values into the lens formula:

1/2.41 = 1/13 - 1/u

Simplifying:

1/2.41 + 1/13 = 1/u

Multiplying both sides by 2.41 x 13:

13 + 2.41 = 2.41 x 13/u

15.41 = 31.33/u

Dividing both sides by 31.33:

u = 31.33/15.41
u ≈ 2.03 cm

Therefore, the lens is approximately 2.03 cm away from the slide.

(b) If the image is virtual:
In this case, the object distance (u) will be negative.

Given:
f = 2.41 cm
v = 13.0 cm

Substituting these values into the lens formula:

1/2.41 = 1/13 - 1/u

Simplifying:

1/2.41 + 1/13 = 1/u

Multiplying both sides by 2.41 x 13:

13 + 2.41 = 2.41 x 13/u

15.41 = 31.33/u

Dividing both sides by 31.33:

u = 31.33/15.41
u ≈ 2.03 cm

In this case, the object distance (u) is also approximately 2.03 cm.

Therefore, the lens is approximately 2.03 cm away from the slide in both cases.

To find the distance between the lens and the slide in both cases, we can use the thin lens equation:

1/f = 1/di - 1/do

where f is the focal length of the lens, di is the image distance, and do is the object distance.

(a) If the image is real, the image distance (di) is positive (since the image is formed on the opposite side of the lens from the object). We are given that di = 13.0 cm and f = 2.41 cm.

Substituting these values into the thin lens equation, we get:

1/2.41 = 1/13 - 1/do

To find the object distance (do), we can rearrange the equation:

1/do = 1/13 - 1/2.41

1/do = 0.077 - 0.414

1/do = -0.337

do = -1/0.337

do ≈ -2.965 cm

Since the object distance must be positive (since the object is on the same side of the lens as the light source), the answer is NONE in this case.

(b) If the image is virtual, the image distance (di) is negative (since the image is formed on the same side of the lens as the object). We are given that di = -13.0 cm and f = 2.41 cm.

Substituting these values into the thin lens equation, we get:

1/2.41 = 1/-13 - 1/do

To find the object distance (do), we can rearrange the equation:

1/do = -1/13 - 1/2.41

1/do = -0.077 - 0.414

1/do = -0.491

do = -1/0.491

do ≈ -2.035 cm

Since the object distance must be positive, the answer is NONE in this case as well.

Therefore, in both cases, there is no real solution for the distance between the lens and the slide.