Lens question

[Addallat]

What is true of the image formed by an object 50 cm from a convex lens that has a radius of curvature equal to 60 cm?

A. It is inverted, real image that is found 75 cm from the lens
B. It is an inverted, real image that is found 30 cm from the lens
C. It is an upright virtual image that is found 30 cm from the lens
D. It is an upright, virtual image that is found 75 cm from the lens


The correct answer is A

if radius of curvature is 60cm, then focal point will be 30cm.
Object farther from the lens than the focal point will be SIR (smaller, inverted, real) = A or B
Now because its smaller, it will be farther from the lens than the object (>30 cm) = A

noticed that you stated the image should be "SIR" small inverted and real when the object is outside the focal length

but from the magnification equation

m = - di/do

m = - (75)/(50)

i get that the image is inverted and real but shouldn't the image be 1.5 times LARGER than the object when you solve for the magnification equation above? did i make another mistake somewhere?

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[Jepstein30]

Can someone please help me with the following question from TBR

What is true of the image formed by an object 50 cm from a convex lens that has a radius of curvature equal to 60 cm?

A. It is inverted, real image that is found 75 cm from the lens
B. It is an inverted, real image that is found 30 cm from the lens
C. It is an upright virtual image that is found 30 cm from the lens
D. It is an upright, virtual image that is found 75 cm from the lens


Why can't i use the lens equation for this question? 1/o+1/i=1/f

radius of curvature is 1/2(60)= 30 cm (positive since it's a converging convex lens)

1/30-1/50 = -1/20 so i = - 20 cm

the image is 20 cm upright virtual image in front of the lens

The correct answer is A

Likely have a problem with the sign notation. Anyways though, try to avoid that equation if possible.

Convex lens = converging

if radius of curvature is 60cm, then focal point will be 30cm.
Object farther from the lens than the focal point will be SIR (smaller, inverted, real) = A or B
Now because its smaller, it will be farther from the lens than the object (>30 cm) = A

EDIT:

1/30 - 1/50 = 5/150 - 3/150 = 2/150

150/2 = 75 cm

You can't subtract fractions by just taking the difference of the denominators..

 

[Addallat]

ah crud math fail in subtracting fractions thank you

 

[Jepstein30]

ah crud math fail in subtracting fractions thank you

Yea, but still try to work conceptually here:

Diverging system = always SUV (smaller, upright, virtual)
Converging system=
Beyond focal point- SIR (smaller, inverted, real)
At focal point- no image
Within focal point- LUV (larger, upright, virtual)

smaller = farther away from lens/mirror
larger = closer

 

[Addallat]

if radius of curvature is 60cm, then focal point will be 30cm.
Object farther from the lens than the focal point will be SIR (smaller, inverted, real) = A or B
Now because its smaller, it will be farther from the lens than the object (>30 cm) = A

Just noticed that you stated the image should be SIR small inverted and real when the object is outside the focal length

but from the magnification equation

m = - di/do

m = - (75)/(50)

i get that the image is inverted and real but 1.5 times LARGER than the object? did i make another mistake somewhere?

 

[Addallat]

..

 

[PlumbLine]

You are correct that the image is larger. It would only be SIR if Do was greater than 2F. Your object is between 1F and 2F, so it is inverted, magnified, real. A Do less than 1F would be upright, virtual, and magnified.

Basically there are 3 "zones" to keep track of with convex lenses, and yours is in the tricky zone of 1F to 2F.

I suck at lenses though so someone else may need to verify. I think the "hyperphysics" web site explains it pretty well.

 

[Jepstein30]

You are correct that the image is larger. It would only be SIR if Do was greater than 2F. Your object is between 1F and 2F, so it is inverted, magnified, real. A Do less than 1F would be upright, virtual, and magnified.

Basically there are 3 "zones" to keep track of with convex lenses, and yours is in the tricky zone of 1F to 2F.

I suck at lenses though so someone else may need to verify. I think the "hyperphysics" web site explains it pretty well.

Oh ****, my bad. don't know why I got that wrong

real rules for converging systems:
beyond R (2F) = SIR (smaller, inverted, real)
at R = SSIR (same size, inverted, real)
between R and F = LIR (larger, inverted real)
at F = no image
between F and lens = LUV (larger, upright, virtual)

basically the image is getting bigger as you move the object closer, which makes sense.

PS PM me next time if you have a followup question, that way I will know to come back ASAP.

 

[Addallat]

you guys are great thank you!