Ray Optics and Optical Instruments

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Question
CBSEENPH12039955

A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the above red, green and blue wavelength are 1.39, 1.44 nd 1.47



The Prism will

  • separate the blue colour part from the red and green colours

  • separate all the three colours from one another

  • not separate the three colours at all

  • separate the red colour part from the green and blue colours

Solution

D.

separate the red colour part from the green and blue colours

For refractive index of a index.
straight mu space equals space fraction numerator 1 over denominator sin space straight i subscript straight c end fraction space equals space fraction numerator 1 over denominator sin space 45 to the power of straight o end fraction space equals space square root of 2

straight mu subscript red space equals space 1.39 space comma space straight mu subscript green space equals space 1.44 space and space straight mu subscript blue space equals space 1.47
left parenthesis straight mu subscript red space equals space 1.39 right parenthesis space less than thin space straight mu comma space straight mu subscript green space end subscript greater than straight u subscript comma space straight mu subscript blue space greater than straight mu
Thus comma space only space red space colour space do space not space suffer space total space internal space reflection.

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Question
CBSEENPH12040121

A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is given by

  • y/2x

  • y/x

  • 2y/x

  • y/x

Solution

A.

y/2x

When the mirror is rotated by θ angle reflected ray will be rotated by 2θ.

y/x = 2θ

θ = y/2x

Question
CBSEENPH12039931

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed on an object of height 2 cm placed 30 cm from the lens?

  • Virtual, upright, height = 0.5 cm

  • Real, invented, height = 4 cm

  • Real, inverted, height = 1 cm

  • Virtual, upright, height = 1 cm

Solution

B.

Real, invented, height = 4 cm

In space general space we space have space assumed space straight mu space equals space 1.5
So comma space straight f space equals space 20 space cm

1 over straight f space equals space 1 over straight v space plus 1 over straight u

1 over 20 space equals space 1 over straight v space plus 1 over 30

1 over straight v space equals space 1 over 20 minus 1 over 30 space equals space 10 over 600

straight v equals space 60 space cm

fraction numerator straight h subscript straight i over denominator straight h subscript straight o space end fraction space equals space 2

straight h subscript straight i space equals space 2 space straight x space vertical line straight h subscript straight o vertical line

straight h subscript straight i space equals space 4 space cm
Here, image is real, inverted, magnified field and height of image is 4 cm

Question
CBSEENPH12039863

A concave mirror of focal length f1 is placed at a distance of d from a convex lens of focal length f2. A beam of light coming from infinity and falling on this convex lens concave mirror combination returns to infinity. The distance d must be equal

  • f1 +f2

  • -f1 +f2

  • 2f1+f2

  • -2f1 +f2

Solution

C.

2f1+f2

d= 2f1 +f2

Question
CBSEENPH12039973

A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meets will moves 5 cm closer to the lens. The focal length of the lens is 

  • -10 cm

  • 20 cm

  • -30 cm

  • 5 cm

Solution

C.

-30 cm

Given u = 10 cm, v = 15 cm
1 over straight f space equals space 1 over straight v minus 1 over straight u

1 over straight f space equals space 1 over 15 minus 1 over 10

1 over straight f space equals space fraction numerator 10 minus 15 over denominator 10 end fraction

1 over straight f space equals space fraction numerator negative 5 over denominator 150 end fraction

straight f space equals space minus space 30

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