Ray Optics and Optical Instruments Test 4

Result

Ray Optics and Optical Instruments Test 4

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Weekly Quiz Competition

Question 1
1 / -0

A ray of light going from denser to rarer medium suffers refraction at a concave surface. Which of the following relations is correct?

A

B

C

D

Solution

Solution :
The correct option is Option A.
Laws of refraction;-
The incident ray,the refracted ray and the normal to the refracting surface at the point of incidence lie in the same plane.
For a given pair of media and for a given colour of light the ration between the sine of angle of incidence to the sine of refraction is a constant.This constant is known as refractive index of the second medium with respect to the first medium.
When a ray of light passes through a glass slab, ∠i,∠r and the normal all lie in the same plane.
When a ray of light passes from one medium to another, here from air to glass or glass to air, the ratio sini / sinr = constant.
Question 2
1 / -0

For a plane surface, the formula becomes

A

B

C

D

Solution

For plane mirror R=infinity
so. (n₂-n₁)/R=0
therefore, n₂/V=n₁/U
Question 3
1 / -0

Consider a curved surface between two different media with refractive indices n₁ = 1 and n₂ = 2. The relation between radius of curvature, image distance and object distance is given by

A

B

C

D

Solution

The relation between radius of curvature, image distance and object distance is given by
n₂/v-n₁/u=n₂-n₁
un₂-vn₂/vu=n₂-n₁/R
R=vu(n2-n2)/un₂-vn₂
=uv(2)/u(2)-v(1)
=[uv/2u-v)
Question 4
1 / -0

In the adjoining figure, SS is a spherical surface separating two media of refractive indices n₁ and n₂ where n₁ > n₂. C is the centre of curvature of the spherical surface. An observer, keeping his eye beyond C in the medium of refractive index n₂ views the refracted image of an object AB placed as shown in the medium of refractive index n₁. The image will be:

A
real, upright and diminished

B
virtual, upright and magnified

C
real, inverted and magnified

D
virtual, upright and diminished

Solution

The image will be virtual, upright and diminished.
Question 5
1 / -0

A thick plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face. Then, x is equal to:

A
-2.5cm

B
2.3cm

C
2cm

D
2.1cm

Solution

A thick Plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face, and we have to determine the value of ' x'
According to the picture, the ray of light gets refracted at the interface between the air and the lens from the object 'p' and 'I' is the refracted image of 'p'
object distance 'u'= BO
Image distance 'v'/'x' = BI
we know
n2/v - n1/u = (n2- n1) / R
or 1/v - 1.5/(-3 )= (1-1.5)/ (-5) [where n2= 1, n1= 1.5 ,u= -3 ,R= -5]
so, 1/v = -6/15
or, v = -2.5
so, x is equal to 2.5 cm