+0

# Not really a math question, more physics, but I'm lost so if someone here could help me out I'd really appreciate it!

0
199
3

Not really a math question, more physics, but I'm lost so if someone here could help me out I'd really appreciate it, it's 00:40 and I'm panicking because I don't understand them oof! It's about the lensformula.

a. An arrow of 2cm length stands about 6cm away from a convex lens (f = 2cm). How long is the image of the arrow?

[  ] A 1 cm  [  ] D 4 cm

[  ] B 2 cm  [  ] E 6 cm

[  ] C 3 cm

b. A windmill on an lcd-screen of a beamer is 3 cm high. On the projectionscreen the windmill is 90 cm high. The lens has a strength of 10 dpt.

What's the distance from the lens to the lcd-screen

[  ] A 0,103 cm  [  ] C 10,0 cm

[  ] B 5,15 cm    [  ] D 10,3 cm

Dec 17, 2018

#1
+770
+1

a)

Height of arrow = 2cm

Distance from lense = 6cm

Focal length = 2cm

Formula: 1/f = 1/v +1/u               Note: f = focal length, v = image of arrow's distance from lens, u = object's distance from lens

notice that the values are inverted because they are on the opposite side of the lens (opposite side of the arrow)

-1/2 = -1/6 +1/v

1/v = -1/2 + 1/6, which simplifies into 1/v = -1/3

v = -3 cm

This means that the height of the image is upside down, but it is still 3 cm.

(If you know physics, the virtual image behind the convex lens turns upside down if the object is behind the focal point of the convex lens.)

Height of image/Height of object = -v /u = 3/6 = 1/2

Height of image / 2 = 1/2                                Note: height of image (virtual reflection) of arrow = h

Multiply by 2 on both sides to get \(\boxed{h = 1cm}\).

The answer to a) is \(\boxed{A}\).

Hope this helps,

- Partial Physician (Partial Mathematician)

Dec 18, 2018
edited by PartialMathematician  Dec 18, 2018
#2
+770
+1

For b), try to follow along with what I did in a). Use the variable "v" to represent the "virtual image's distance from the convex lens".

I'll give you the starting variables.

Height of windmill = 3cm = height of object

Height of virtual image = 90 cm = h

Lens strength = 10 dpt = focal length 10 cm

- Partial Physician/Mathematician

PartialMathematician  Dec 18, 2018
#3
0

Height of windmill = 3cm = height of object

Height of virtual image = 90 cm = h

Lens strength = 10 dpt = focal length 10 cm

Formula: 1/f = 1/v +1/u

1/10 = 1/3 + 1/u

1/u = 1/f - 1/u

1/u = 1/10 - 1/3

1/u = -0,2333 or - 7/30

-v /u = - 7/30 / 1/3 = -0,1

so it'd be C then?

Guest Dec 23, 2018