Images from Spherical Mirrors — Question 2

Section 34-4 · Physics pages 5614–5615

Question 2 An object inside the focal length produces an enlarged virtual mirror image Use the paraxial mirror relations 1/p + 1/q = 1/f and m = −q/p, with p positive for a real object, q positive for a real image, and f positive for a concave mirror. A concave mirror has f=15 cm and an object at p=10 cm. The object height is 4 mm. F C V O I Schematic; labels, not drawing lengths, specify the data. (a) Find q and m and classify the image. (b) Find its height and the physical distance between object and image.
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Question 2 - Solution An object inside the focal length produces an enlarged virtual mirror image F C V O I (a) Method and calculation q = 15(10)/(10 −15) = −30 cm . m = −q/p = +3 ⇒virtual, upright, enlarged. (b) Interpretation and further calculation hI = 3(4) = 12 mm . dOI = p + |q| = 40 cm . Physical check The object and virtual image are on opposite sides of the mirror, so their physical separation is the sum of distance magnitudes.

Question 2

An object inside the focal length produces an enlarged virtual mirror image

Use the paraxial mirror relations 1/p+1/q=1/f1/p+1/q=1/f and m=−q/pm=-q/p, with p positive for a real object, q positive for a real image, and f positive for a concave mirror. A concave mirror has f=15 cm and an object at p=10 cm. The object height is 4 mm.

Diagram shown in the original SVG worksheet.

  1. Find q and m and classify the image.

  2. Find its height and the physical distance between object and image.

Question 2 - Solution

An object inside the focal length produces an enlarged virtual mirror image

Diagram shown in the original SVG worksheet.

(a) Method and calculation

q=15(10)/(10−15)=−30cm.q=15(10)/(10-15)=\boxed{-30\,\mathrm{cm}}.m=−q/p=+3⇒virtual, upright, enlarged.\boxed{m=-q/p=+3}\Rightarrow\text{virtual, upright, enlarged}.

(b) Interpretation and further calculation

hI=3(4)=12mm.\boxed{h_I=3(4)=12\,\mathrm{mm}}.dOI=p+|q|=40cm.\boxed{d_{OI}=p+|q|=40\,\mathrm{cm}}.

Physical check

The object and virtual image are on opposite sides of the mirror, so their physical separation is the sum of distance magnitudes.