A fly of height H sits in front of lens 1 on the central axis. the lens forms an

Question

A fly of height H sits in front of lens 1 on the central axis. the lens forms an image of the fly at a distance d = 25 cm from the fly. The image has the same orientation as the fly and a height = 0.3 H. a) What is the object distance (distance from the lens to the fly)? b) What is the focal length of the lens? is it a convex / concave lens? Explain your answer. c) If the object (fly) moved to a distance, 30 cm away from the lens, i) Find the image distance. ii) Find the orientation of the image. iii) Find the magnification of image. iv) Explain the type of the image - Real / virtual.

Explanation / Answer

M = 0.3....SO it is clearly a Concave Lens...Because only it can produce virtual diminished images...

So, v / u = 0.3
SO, v = 0.3 u

Now, |u|-|v| = 25 cm
So, 0.7 u = 25 cm
So, u = 35.71 cm
v = 10.71 cm

Following sign conventions for concave lens :
u = -35.71 cm
v = -10.71 cm
So, 1 / f = 1/ v - 1 / u
So, f = -15.298 cm

c)
u = -30 cm
f = -15.298 cm
Now, i / v = 1 / f + 1 / u

SO, v = -10.13 cm
M = v / u = 0.33
Since it is positive...SO image is ERECT and VIRTUAL

Answers :
a) u = -35.71 cm...you can take positive sign as well

b) f = -15.298 cm
it is clearly a Concave Lens...Because only it can produce virtual diminished images...

c) i) v = -10.13 cm....again u can neglect the sign
ii) Erect...same as that of fly
iii) M = 0.337
iv) Virtual....Because it is on same side as the fly

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