Light and Optics from TBR: How do you figure out image distance?

[Ash366]

Example 10.7b. What is true of image formed by a convex mirror with a focal length of 20cm by an object that is 30cm from the mirror?

These are what I can immediately figure out in my head: I (eye) am positive real is inverted (EK method) indicating it is positive focal distance, Real and inverted. Then Radius of curvature= 40. Thus Radius of curvature < Object distance > focal distance


Im not sure how to accurately figure out image distance without actually putting in the values into lens makers equation. Solution says "If f=25cm then R=50cm, so the image must form beyond 50. How do you easily figure that out without inserting values into the equation?

Then theres the next one 10.7b. where I couldnt get the answer without inserting the values in. Please help thank you!

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[AwayFromReality]

Are you sure the question didn't state a convex lens? Because a convex mirror is a diverging mirror which always forms a Small Upright Virtual image close to the mirror. A convex lens on the other hand is a converging lens which we can solve using the explanation you posted.

So assuming it is convex lens (converging lens) we can use certain rules. These rules were listed in TBR which is what I used to study material.

if the object is located between R and f then the image will be located past R inverted/real/larger.

If the object is located past R then the image will be located between R and f inverted/real/smaller

If the object is located at R then the image will also be located at R inverted/real/same size

If the object is located at f then the image will be non existent.

If the object is located inside of the focal length, the image will be located past the focal length virual/upright/larger

These rules apply to both mirror and lenses as long as both of them are converging. Like I mentioned above converging lens = convex. Converging mirror = Concave.

For diverging lens OR mirror the image is always small upright virtual and inside the focal length.

I memorized these rules for the MCAT. Obviously you don't have to memorize them you can always do the math if the numbers are given or draw the ray diagrams for each situation but that requires a lot of time on the MCAT and memorizing seems to be the easier way

In 10.7b Focal length is 20 therefore radius of curvature is 40. Object is located at 30 ( between R and f) which means image will be located past R (past 40 cm).

 

[Ash366]

Thank you for your great detailed information. I have the 2011 version of TBR, is it supposed to say lens? All the questions are for a converging mirror?

And 10.7b is less than focal point. (less than 20cm). Its confusing me now.

 

[AwayFromReality]

Thank you for your great detailed information. I have the 2011 version of TBR, is it supposed to say lens? All the questions are for a converging mirror?

And 10.7b is less than focal point. (less than 20cm). Its confusing me now.

Just looked up 10.7b in my book (2013 edition) and you're right it is convex mirror. In this case we simply ignore the converging rules and focus on the diverging rule (diverging only has one rule). Since it is a convex mirror, it is diverging which means the rule for diverging lens or mirror still applies. The image in this case will be small upright virtual and inside the focal length. For a diverging mirror or lens it does not matter where the object is you will always get small upright virtual image inside the focal length

I was just confused by your original post and thought you were referring to converging rather than diverging. You seem to be mentioning both focal lengths of 25 and 20 but you only mentioned one problem (10.7b) Now that I looked up 10.7a I realize that was the question that was dealing with 25cm focal length.

10.7a is dealing with a converging mirror with a focal length of 25 therefore R of 50cm. Object is located at 40 cm (between f and R). According to the rules above for a converging mirror the image will be past R (past 50cm) real and inverted. (Also, side note, I noticed that for the explanation of 10.7a TBR wrote "...converging mirror form upright, real images, eliminating choice D". There is a typo here it should not say upright but rather inverted. Real images are always inverted and Virtual images are always upright)

Does this help solve your confusion?