1. a concave mirror has a focal length of 40cm has 2 cm tall object placed 45 cm

Question

1. a concave mirror has a focal length of 40cm has 2 cm tall object placed 45 cm in front of it. the image formed is
2. A light Ray in air enters water at an angle of incidence of 40. Water has an index of reflection of 1.33. The angle of refraction in the water is
3. A lens of focal length 20 cm has an object of height 3 cm placed 100 cm in front of the image formed is

4. If light of wavelength 600 mm in air enters a medium with index of refraction wavelength in this new medium is


5. Two narrow slits spaced 50 microns apart are exposed to light of 600 nm. At what angle does the first minimum ( dark space ) occur in the interference pattern 1. a concave mirror has a focal length of 40cm has 2 cm tall object placed 45 cm in front of it. the image formed is
2. A light Ray in air enters water at an angle of incidence of 40. Water has an index of reflection of 1.33. The angle of refraction in the water is
3. A lens of focal length 20 cm has an object of height 3 cm placed 100 cm in front of the image formed is

4. If light of wavelength 600 mm in air enters a medium with index of refraction wavelength in this new medium is


5. Two narrow slits spaced 50 microns apart are exposed to light of 600 nm. At what angle does the first minimum ( dark space ) occur in the interference pattern 1. a concave mirror has a focal length of 40cm has 2 cm tall object placed 45 cm in front of it. the image formed is
2. A light Ray in air enters water at an angle of incidence of 40. Water has an index of reflection of 1.33. The angle of refraction in the water is
3. A lens of focal length 20 cm has an object of height 3 cm placed 100 cm in front of the image formed is

4. If light of wavelength 600 mm in air enters a medium with index of refraction wavelength in this new medium is


5. Two narrow slits spaced 50 microns apart are exposed to light of 600 nm. At what angle does the first minimum ( dark space ) occur in the interference pattern

Explanation / Answer

1. 1/f = 1/do + 1/di

ho = 2 cm do = 45 cm f = 40 cm

1/40 = 1/45 +1/di

di = 360 cm

hi/ho = - di/do

hi/2 = -360/45

hi = 16 cm

2. 1*sin (40) = 1.33 sin(theta r)

sin (theta r ) = 0.56 so theta r =34 degree

3.1/u+1/v = 1/f

1/100 +1/v =1/20

v = distance of image = 25 cm

hi/ho = v/u = 25/100

hi = 3*25/100 = 3/4 cm = 0.75 cm

4.. wavelength (water) = (air) /(4/3) = 600*3/4 = 450 mm

5.
d sin = ½ , sin ~ = ½ /d = 600*10-9/ 50*2*10-6
= 6*10-3 radians

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