Suppose W is the power of the lamp. Then amount of energy needed to print photo is E=2.5*W/4??602 Now suppose t is the required time when the lamp is kept at 1.5 m away. So we have E=W/4??1202t Equating energy for both we get t=10 s.
46) To print a photograph from a negative, the time of exposure to light from a lamp placed 60 cm away is 2.5 s. What exposure time is required if the lamp is placed 1.2 m away?
Answer is:
10 s
Explanation:
Related Optics MCQ with Answers
Answer is:
30 cm
Explanation:
n1/u +n2/v=(n2-n1)/r for first refraction by convex surface: u??, v=rn2/(n2-n1) for plane reflection: r is infinite,v'=-v for second refraction: 1.5/v' -1/v''=(1-1.5)/r 1/v''+1/v=1/f power=1/f
Answer is:
1.5
Explanation:
n1/u +n2/v=(n2-n1)/r for first refraction by convex surface: u??, v=rn2/(n2-n1) for plane reflection: r is infinite,v'=-v for second refraction: 1.5/v' -1/v''=(1-1.5)/r 1/v''+1/v=1/f power=1/f
Answer is:
complete image will be formed
Explanation:
A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen complete image will be formed
Answer is:
its wavelength decreases
Explanation:
When a ray of light enters a glass slab from air its wavelength decreases