In a projection lantern of focal length f, the object distance u, is such that
u > 2f > f
u < f < 2f
u = f < 2f
f < u < 2f
Correct answer is D
No explanation has been provided for this answer.
5cm
10cm
30cm
40cm
Correct answer is B
V = -2u
F = \(\frac{uv}{u + v}\)
20 = \(\frac{uv - 2u}{u - 2u}\)
u = 10cm
4.0 x 1014Hz
6.0 x 1014Hz
7.5 x 1014Hz
9.0 x 1014Hz
Correct answer is A
1.5 = \(\frac{6 \times 10^{14}}{F}\)
F = 4 x 1014Hz
the focal length
two times the focal length
the distance of the image
two times the radius of curvature
Correct answer is A
No explanation has been provided for this answer.
200Hz
400Hz
800Hz
1600Hz
Correct answer is B
F = \(\frac {1}{2L}\) \(\sqrt\frac{T}{m}\)
T2 = 4T1 ; L2 = 2L1
\(\frac{2F_1L_1}{2F_2L_2}\) = \(\sqrt\frac{T_1}{T_2}\)
\(\frac{400\times L_1}{F \times L_1}\) = \(\sqrt\frac{T_1}{4T_1}\)
F = 400Hz