A thin lens of focal length f and its aperture diameter d, forms a real image of intensity I. Now the central part of the aperture of diameter (d2) is blocked by an opaque paper. The focal length and image intensity would change to
A
f2,I2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
f,I4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3f4,I2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
f,3I4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is Df,3I4 On blocking the central part of the lens, focal length does not change. Intensity of the image is directly proportional to the area of the lens. Initial area= A1=π(d2)2=πd24 Final area= A2=π(d2)2−π(d4)2=πd24−πd216=3πd216 Hence I2I1=A2A1=34 I2=34I1=34I