another name of night blindness

MCQ Questions for Class 12 Medical Physics Ray Optics And Optical Instruments Quiz 4

If the image of distant objects is formed in front of the retina, the defect of vision may be:

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Myopia
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Hypermetropia
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Cataract
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Astigmatism

A plane mirror produces an image that is

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Real, inverted and larger than the object.
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Real, upright and same size as the object.
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Real upright and smaller than the object.
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Virtual, upright and the same size of the object.

One cannot see through fog because:

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fog absorbs light
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light is scattered away by the droplets in fog
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light suffers total internal reflection at the droplets in fog
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the refractive index of fog is infinity

The final image formed by an astronomical telescope is:

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real and erect.
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virtual and erect.
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real and inverted.
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virtual and inverted.

Formula of focal length in convex lens is

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$$\displaystyle f = \frac{u+v}{u-v}$$
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$$\displaystyle f = \frac{u\times v}{u-v}$$
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$$\displaystyle f = \frac{u-v}{u+v}$$
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$$\displaystyle f = \frac{u+v}{u+v}$$

A person is suffering from myopia. Then :

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he can not see anything.
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he cannot see distant objects clearly.
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he can see only distant objects.
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he can not see nearly objects.

Which of the following options correctly describe respective image?

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A short-sighted eye corrected by a convex lens.
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A short sighted eye corrected by concave lens.
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A healthy eye.
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A short sighted eye corrected by convex lens.

Amount of light entering in the eye is controlled by:

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Pupil
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Iris
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Cornea
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Eye lens

The focal length of a lens is 0.1 m. Then the lens must be :

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Convex
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Concave
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Cylindrical
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None of these

Long-sightedness is caused by the eyeball being too short. It can be corrected by the use of a:

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convergent lens
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plane mirror
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divergent lens
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None of these

The bending of light as it passes from one medium into another is commonly known as

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Reflection
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Refraction
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Scattering
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Dispersion

Which of the following lens is used to correct myopia?

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Convex lens
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Concave lens
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Cylindrical lens
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None of these

When light travels from one medium into another it suffers

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Reflection
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Refraction
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Dispersion
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None of these

The figure shows cross section of a hollow glass tube of internal radius $$r$$, external radius $$R$$ and index of refraction $$n$$. For two rays DE and ABC (in which DE lies on ODE and DE is parallel to BC), the separation $$r_1$$ will be :

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$$r_1 = (n - 1) R$$
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$$r_1 = n^2 R$$
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$$r_1 = nr$$
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$$r_1 = n^2 r$$

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

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separate part of the red color from the green and blue colors.
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separate part of the blue color from the red and green colors.
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Separate all the three colors from the other two colors.
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not separate even partially any color from the other two colors.

In case of hypemetropia

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The image of near objects is formed in front of retina
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The image of near objects is formed behind the retina
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A concave lens should be used for correction
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A convex lens cannot be used for correction

Why are red coloured lights are used in traffic signals to stop the vehicles?

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minimum wavelength and less scatter.
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maximum wavelength and more scatter.
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maximum wavelength and less scatter.
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less wavelength and less scatter.

A thin lens with focal length f to be used as magnifying glass. Which of the following statements regarding the situation is true?

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A converging lens may be used, and the object be placed at a distance greater than 2f from the lens.
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A diverging lens may be used, and the object be placed between f and 2f from the lens.
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A converging lens may be used, and the object be placed at a distance less than f from the lens.
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A diverging lens may be used, and the object be placed at any point other than the focal point.

The index of refraction of diamond is 2.The velocity of light in diamond in cm/s is -

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$$6\times 10^{10}$$
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$$2\times 10^{10}$$
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$$3\times 10^{10}$$
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$$1.5\times 10^{10}$$

A ray of light is incident normally on the first refracting face of the isosceles prism of refracting angle A. The ray of light comes out at grazing emergence. If one half of the prism (shaded position) is knocked off, the same ray will :

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emerge at an angle of emergence $$sin^{-1} \left ( \frac{1}{2} sec A/2 \right )$$
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not emerge out of the prism
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emerge at an angle of emergence $$sin^{-1} \left ( \frac{1}{2} sec A/4 \right )$$
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None of these

Explanation

Given: A ray of light is incident normally on the first refracting face of the isosceles prism of refracting angle A. The ray of light comes out at grazing emergence.

To find the angle emergence angle of the same if one half of the prism is knocked off

Solution:

Case (i): when ray of light is incident normally on first refracting face of the isosceles prism. (refer fig(i))

Let the ray X is incident normally on first refracting face, AB of the isosceles triangle ABC.

As it it given refracting angle, $$\angle {XYZ} = A$$

Hence $$\angle XYA = (90-A)$$

In right angled traingle $$AOY$$

$$180^\circ = \angle AOY + \angle OYA +\angle YAO\\\implies \angle YAO=180-90-(90-A)=A.........(i)$$

And as the comes out at grazing emergence hence angle of emergence in this case is $$90^\circ$$

So we know,

$$\mu_1\times \sin i=\mu_2\times \sin e........(ii)$$.

Here $$\mu_1=\mu_p$$ refractive index of prism,

$$\mu_2=\mu_a=1$$ as it is refractive index of air,

angle of incidence $$i=A$$ (given) and

angle of emergence $$e=90^\circ$$ (given).

Therefore, by substituting these values in equation (ii), we get

$$\mu_p\times \sin A=1\times \sin 90\\\implies \mu_p=\dfrac 1{\sin A}.........(iii)$$

Case (ii): when half of the prism is knocked off (refer fig(ii))

Then $$\angle YAO=\dfrac A2$$

In right-angled triangle,

$$180^\circ=\angle AOY + \angle OYA +\angle YAO\\\implies \angle OYA=180-90-\left(\dfrac A2\right)\\\implies \angle OYA=90-\dfrac A2$$

And angle of incidence, $$\angle XYZ=i=90-\left(90-\dfrac A2\right)=\dfrac A2$$

So the equation (ii), in this case will become

$$\mu_p\times \sin \left(\dfrac A2\right)=\mu_a\times \sin e\\\implies \dfrac 1{\sin A}\times \sin \left(\dfrac A2\right)=\sin e$$

$$\sin e = \dfrac {\sin \left(\dfrac A2\right)}{2\sin \left(\dfrac A2\right)\cos \left(\dfrac A2\right)}$$ [$$\because \sin A=2\sin \left(\dfrac A2\right)\cos \left(\dfrac A2\right)$$]

Therefore,

$$\sin e=\dfrac{\sec \left(\dfrac A2\right)}2$$ [$$\because \sec A=\dfrac 1{\cos A}$$]

Hence incident ray will emerge at angle of emergence

$$e=\sin^{-1} \left\{\dfrac 12 \sec \left(\dfrac A2\right)\right\}$$

The image formed by the eye lens on the retina is :

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Real, upright and enlarged
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Real, upright and diminished
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Real, inverted and diminished
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Virtual, inverted and diminished

A terrestrial telescope is made by introducing an erecting lens of focal length, $$f$$, between the objective and eyepiece lens of an astronomical telescope. This causes the length of telescope tube to increase by an amount equal to

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$$f$$
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$$2f$$
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$$3f$$
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$$4f$$

The cause of mirage formation in desert areas is

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The refractive index of atmosphere decreases with decrease in height
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The refractive index of atmosphere increases with decrease in height
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The refractive index of atmosphere does not change with height
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Scattering

If the light moving in a straight line bends by small but fixed angle while entering in different medium , it may be a case of :

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reflection of light
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refraction of light
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scattering of light
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dispersion of light

An astronomical telescope and a Galilean telescope use identical objective lenses. They have the same magnification, when both are in normal adjustment. The eyepiece of the astronomical telescope has a focal length f.

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The tube length of the two telescope differ by f.
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The tube length of the two telescopes differ by 2f.
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The Galileans telescope has shorter tube length.
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The Galileans telescope has longer tube length.

When objects at different distance are seen by the eye, which of the following remain constant?

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The focal length of the eye-lens.
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The object- distance from the eye-lens.
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The radii of curvature of the eye-lens.
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The image-distance from the eye-lens.

The distance of the eye-lens from the retina is $$x$$. For a normal eye, the maximum focal length of the eye-lens

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$$= x$$
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$$< x$$
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$$> x$$
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$$= 2x$$

in which of the following is the final image erect?

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Simple microscope.
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Compound microscope
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Astronomical microscope
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Galilean telescope.

A ray of light incident normally on the first face of an equilateral prism undergoes total internal reflection at the second face if refractive index $$ \mu $$ is

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less than $$ \dfrac { \sqrt { 3 } }{ 2 } $$
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less than $$ \dfrac { 2 }{ \sqrt { 3 } } $$
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more than $$ \dfrac { \sqrt { 3 } }{ 2 } $$
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more than $$ \dfrac { 2 }{ \sqrt { 3 } } $$

Identify the wrong description of the given figures.

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1 represents far sightedness
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2 is correction for short sightedness
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3 represent far sightedness
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4 represents correction for far sightedness

A ray of light falls on the surface of spherical glass paperweight making an angle $$\alpha$$ with the normal and is refracted in the medium at an angle $$\beta$$. The angle of deviation of the emergent ray from the incident ray is

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$$\left( \alpha -\beta \right) $$
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$$ 2\left( \alpha -\beta \right) $$
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$$ \dfrac { \left( \alpha -\beta \right) }{ 2 } $$
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$$ \left( \alpha +\beta \right) $$

A ray of light enters a spherical drop of water of refractive index $$\mu$$ as shown in fig. Select the correct statement(s).

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Incident rays are partially reflected at point A
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Incident rays are totally reflected at point A
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Incident rays are totally transmitted through A
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None of these

The refractive index of a prism is $$2$$. The prism can have a maximum refracting angle of

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$$90^{\circ}$$
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$$60^{\circ}$$
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$$45^{\circ}$$
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$$30^{\circ}$$

A light ray hits the pole of a thin biconvex lens as shown in fig. The angle made by the emergent ray with the optic axis will be approximately

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$$0^o$$
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$$(1/3)^o$$
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$$(2/3)^o$$
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$$2^o$$

A given ray of light suffers minimum deviation in an equilateral prism $$P$$. Additional prism $$Q$$ and $$R$$ of identical shape and of the same material as $$P$$ are now added as shown in the figure. The ray will now suffer.

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Greater deviation
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No deviation
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Same deviation as before
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Total internal reflection

The relation among u, v and f for a mirror is

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$$f = uv/(u+v)$$
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$$v = fu/(u+f)$$
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$$u = fv/(f+v)$$
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All of these

In an astronomical telescope, the distance between the objective and the eyepiece is $$36\ cm$$ and the final image is formed at infinity. The focal length $$f_{0}$$ of the objective and the focal length $$f_{e}$$ of the eyepiece are

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$$f_{0} = 45\ cm$$ and $$f_{e} = -9\ cm$$
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$$f_{0} = 50\ cm$$ and $$f_{e} = 10\ cm$$
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$$f_{0} = 7.2\ cm$$ and $$f_{e} = 5\ cm$$
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$$f_{0} = 30\ cm$$ and $$f_{e} = 6\ cm$$

Hypermetropia is

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same thing as shortsightedness
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same thing as long sightedness
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sane as cataract
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another name of night blindness

The eye forms a real and inverted image of an object on the retina. If the eye lens is defective, the image of objects may be formed in front of or behind the retina. Persons in whom the image is formed in front of the retina are not able to see the distant objects clearly while if the image is formed behind the retina the person is not able to see a near object clearly. The latter kind of people is said to be longsighted.
Persons suffering from myopia cannot see clearly the

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Distant objects
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Near objects
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Spherical objects
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Neither near nor distant

A fish is looking at a $$1.0m$$ high plant of the edge of the pond. Will the plant appear shorter or taller than its actual height, to the fish

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taller
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shorter
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no change
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can't say

What characteristic property of light is responsible for the blue colour of the sky?

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Interference
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Diffraction
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Scattering
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Dispersion

The device used to observe stars and other heavenly bodies is

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periscope
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microscope
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telescope
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compound microscope

An astronaut in a spaceship the sky appears black due to:

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Absence of atmosphere in his neighbourhood
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Light from the sky is absorbed by the medium surrounding him
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The fact at the height, sky radiations are only in the infra-red and the ultraviolet region
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None of these

A ray of light $$OP$$ passes through a right angled prism as shown in the adjacent diagram. State the angles of incidence at the faces $$AC$$ and $$BC$$. Given critical angle is $$42^o$$.

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Angle of incidence at the face $$AC=45^o$$ and angle of incidence at the face $$BC=90^o$$
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Angle of incidence at the face $$AC=90^o$$ and angle of incidence at the face $$BC=45^o$$
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Angle of incidence at the face $$AC=45^o$$ and angle of incidence at the face $$BC=0^o$$
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Angle of incidence at the face $$AC=0^o$$ and angle of incidence at the face $$BC=45^o$$