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Question
A converging mirror M1, a point source S and a diverging mirror M2 are arranged as shown in fig. The source is placed at a distance of 30 cm from M1. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. Find the distance between the two mirrors.
A converging mirror M1, a point source S and a diverging mirror M2 are arranged as shown in fig. The source is placed at a distance of 30 cm from M1. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. Find the distance between the two mirrors.
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Solution
The correct option is C 50 cm
Let's draw the required ray diagram to check how the two reflections may take place.
We know that for a real object (which gives diverging rays) convex mirrors always form a virtual image behind them. So, the convex mirror has to be placed such that the rays falling on it are converging, only then after the second reflection from the convex mirror the image can be formed on the object itself. So, in our diagram, the mirrors are placed such that had M2 not been there, M1 would have made the image at S′. (After the first reflection) Let's assume distance between the mirrors to be d. Let's find the position of S′. Applying mirror formula for M1,
1v=1f−1u=−120+130=−160 v = 60 cm (Hence S′ is at a distance of 60 cm from the concave mirror)
Now, a very interesting thing to note here is that for the convex mirror (M2), the rays that are being incident , appear to be meeting at S′, so S′
serves as an (virtual) object for M2. Also the object distance forM2 is in that condition positive. (Notice these two facts in the diagram) Also, M2 forms the image on the object itself i.e. S.
Hence for M2 from the diagram we can write
u = 60 − d
v = − (d −30)
f = + 20
Now, Applying the mirror formula for M2,
1v+1u=1fWhich gives us130−d+160−d=120
Which on solving yields the following quadratic equation :
d2−50d=0d(d−50)=0Which gives,d=50 cm.
Let's draw the required ray diagram to check how the two reflections may take place.
We know that for a real object (which gives diverging rays) convex mirrors always form a virtual image behind them. So, the convex mirror has to be placed such that the rays falling on it are converging, only then after the second reflection from the convex mirror the image can be formed on the object itself. So, in our diagram, the mirrors are placed such that had M2 not been there, M1 would have made the image at S′. (After the first reflection) Let's assume distance between the mirrors to be d. Let's find the position of S′. Applying mirror formula for M1,
1v=1f−1u=−120+130=−160
v = 60 cm (Hence S′ is at a distance of 60 cm from the concave mirror)
Now, a very interesting thing to note here is that for the convex mirror (M2), the rays that are being incident , appear to be meeting at S′, so S′serves as an (virtual) object for M2. Also the object distance forM2 is in that condition positive. (Notice these two facts in the diagram)
Also, M2 forms the image on the object itself i.e. S.
Hence for M2 from the diagram we can write
u = 60 − d
v = − (d −30)
f = + 20
Now, Applying the mirror formula for M2,
1v+1u=1fWhich gives us130−d+160−d=120
Which on solving yields the following quadratic equation :
d2−50d=0d(d−50)=0Which gives,d=50 cm.
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