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  1. Problems
  2. /A concave mirror has a focal length of 14 cm…
MediumPhysicsHigh School (9–10)Optics › Reflection of LightNumericalBoard Exam

A concave mirror has a focal length of 14 cm…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A concave mirror has a focal length of 14 cm14\ \text{cm}14 cm. Find the image distance and the magnification for an object at 56 cm56\ \text{cm}56 cm, 28 cm28\ \text{cm}28 cm, 21 cm21\ \text{cm}21 cm and 10.5 cm10.5\ \text{cm}10.5 cm in front of the mirror. Say what the image is like each time.

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cm
cm
cm
Need a reminder? The key ideaShowHide

New Cartesian signs: distances from the pole, against the incoming light negative. So uuu is negative, a concave mirror has a negative fff, and a convex mirror a positive fff. Mirror formula 1v+1u=1f\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}v1​+u1​=f1​; magnification m=h′h=−vum = \dfrac{h'}{h} = -\dfrac{v}{u}m=hh′​=−uv​. A negative mmm means real and inverted.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. u=−56 cmu = -56\ \text{cm}u=−56 cm. 1v=−114 cm+156 cm=−356 cm\dfrac{1}{v} = -\dfrac{1}{14\ \text{cm}} + \dfrac{1}{56\ \text{cm}} = -\dfrac{3}{56\ \text{cm}}v1​=−14 cm1​+56 cm1​=−56 cm3​, so v=−18.7 cmv = -18.7\ \text{cm}v=−18.7 cm and m=−0.33m = -0.33m=−0.33. Real, inverted, smaller.

Step 2. u=−28 cmu = -28\ \text{cm}u=−28 cm. The object is at CCC: v=−28 cmv = -28\ \text{cm}v=−28 cm and m=−1m = -1m=−1. Real, inverted, the same size.

Step 3. u=−21 cmu = -21\ \text{cm}u=−21 cm. 1v=−114 cm+121 cm=−142 cm\dfrac{1}{v} = -\dfrac{1}{14\ \text{cm}} + \dfrac{1}{21\ \text{cm}} = -\dfrac{1}{42\ \text{cm}}v1​=−14 cm1​+21 cm1​=−42 cm1​, so v=−42 cmv = -42\ \text{cm}v=−42 cm and m=−2m = -2m=−2. Real, inverted, larger.

Step 4. u=−10.5 cmu = -10.5\ \text{cm}u=−10.5 cm. 1v=−114 cm+110.5 cm=142 cm\dfrac{1}{v} = -\dfrac{1}{14\ \text{cm}} + \dfrac{1}{10.5\ \text{cm}} = \dfrac{1}{42\ \text{cm}}v1​=−14 cm1​+10.5 cm1​=42 cm1​, so v=+42 cmv = +42\ \text{cm}v=+42 cm and m=+4m = +4m=+4. Virtual, upright, larger.

Answer: v=−18.7 cmv = -18.7\ \text{cm}v=−18.7 cm, −28 cm-28\ \text{cm}−28 cm, −42 cm-42\ \text{cm}−42 cm and +42 cm+42\ \text{cm}+42 cm; m=−0.33m = -0.33m=−0.33, −1-1−1, −2-2−2 and +4+4+4.

What do we see? As the object comes closer, the real image moves out and grows. Inside the focus, it jumps behind the mirror.

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