JEE Advanced Test- 10 - JEE MCQ

A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The object is at 15 cm from the lens. The length of the image is:

An object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between object and its image will be :

A ray of light falls on a plane mirror. When the mirror is turned, about an axis which is at right angle to the plane of the mirror through 20º the angle between the incident ray and new reflected ray is 45º. The angle between the incident ray and original reflected ray was therefore : 

The focal length of a lens of refractive index 3/2 is 10 cm in air. The focal length of that lens in a medium of refractive index 7/5 is:

A converging lens of focal length f is placed just above a water surface parallel to the surface without touching it. A point source of light S is placed inside the water vertically before the lens at a depth f from it. This arrangement will produce:

A small air bubble is trapped inside a transparent cube of size 12 cm. When viewed from one of the vertical faces, the bubble appears to be at 5 cm from it. When viewed from opposite face, it appears at 3 cm from it.

A parallel beam of light (l =5000 Å) is incident at an angle a = 30° with the normal to the slit plane in a young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is I₀. Point O is equidistant from S₁ & S₂.The distance between slits is 1mm.    

Figure shows two thin slabs of glass, one is rectangular and the other is triangular wedge shaped. A monochromatic light incident nearly normally on the slabs as shown in figure.

A Young’s double slit experiment is conducted in water (m₁) as shown in the figure and a glass plate of thickness t and refractive index m2 is placed in the path of S₂. The magnitude of the phase difference at O is (Assume that ‘l’ is the wavelength of light in air). O is symmetrical with respect to S₁ and S₂ .

Statement-1 : Two coherent point sources of light having nonzero phase difference are separated by small distance. Then on the perpendicular bisector of line segment joining both the point sources,  constructive interference cannot be obtained.

Statement-2 : For two waves from coherent point sources to interfere constructively at a point, the magnitude of their phase difference at that point must be 2mp ( where m is a nonnegative integer) .

Statement-1 : A spherical surface of radius of curvature R separates two media of refractive index n₁ and n₂ as shown. If an object O (a thin small rod) is placed upright on principal axis at a distance R from pole (i.e, placed at centre of curvature), then the size of image is same as size of object.

Statement-2 : If a point object is placed at centre of curvature of spherical surface separating two media of different refractive index, then the image is also formed at centre of curvature, i.e., image distance is equal to object distance.

Statement-1 : A parallel beam of light is incident on a thin convex lens and is also parallel to the principal axis of convex lens as shown. The magnitude of deviation of each ray of this beam produced by given convex lens is different.

Statement-2 : A thin convex lens can be assumed to be made of prisms of small angles. The magnitude of deviation produced by prism of small angle for small angles of incidence depends on angle of prism. Each ray of the beam in situation of statement-1 is incident on a prism of different angles, hence the magnitude of deviation for each ray is different.

Statement-1 : A ray is incident from outside on a glass sphere surrounded by air as shown. This ray may suffer total internal reflection at second interface.

Statement 2 : For a ray going from denser to rarer medium, the ray may suffer total internal reflection.

In the figure an arrangement of young's double slit experiment is shown. A parallel beam of light of wavelength 'l' (in medium n₁) is incident at an angle 'q' as shown. Distance S₁O = S₂O. Point 'O' is the origin of the coordinate system. The medium on the left and right side of the plane of slits has refractive index n1 and n2 respectively. Distance between the slits is d. The distance between the screen and the plane of slits is

D. Using D = 1m, d = 1mm, q = 30°, l = 0.3mm, n₁ = 4/3, n₂ = ,   10/9

answer the following :

Q. The y-coordinate of the point where the total phase difference between the interefering waves is zero, is 

In the figure an arrangement of young's double slit experiment is shown. A parallel beam of light of wavelength 'l' (in medium n₁) is incident at an angle 'q' as shown. Distance S₁O = S₂O. Point 'O' is the origin of the coordinate system. The medium on the left and right side of the plane of slits has refractive index n1 and n2 respectively. Distance between the slits is d. The distance between the screen and the plane of slits is

D. Using D = 1m, d = 1mm, q = 30°, l = 0.3mm, n₁ = 4/3, n₂ = 10/9,     
answer the following :

Q. If the intensity due to each light wave at point 'O' is I₀ then the resultant intensity at point 'O' will be - 

In the figure an arrangement of young's double slit experiment is shown. A parallel beam of light of wavelength 'l' (in medium n₁) is incident at an angle 'q' as shown. Distance S₁O = S₂O. Point 'O' is the origin of the coordinate system. The medium on the left and right side of the plane of slits has refractive index n₁ and n₂ respectively. Distance between the slits is d. The distance between the screen and the plane of slits is

D. Using D = 1m, d = 1mm, q = 30°, l = 0.3mm, n₁ = 4/3, n₂ = 10/9,     
answer the following :

Q. y-coordinate of the nearest maxima above 'O' will be - 

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.

A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.

A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 

Q. Minimum focal length of eye lens of a normal person is

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.

A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.

A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 

Q. Maximum focal length of eye lens of normal person is

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.

A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.

A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.

The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.

For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 

Q. A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be.

Match the following :         
             Column-I                                                    Column-II
    (A)     Object is between optic center and 1st        (p) Image is inverted
             principle focus in a diverging lens
    (B)    Object is between optic center and 1st         (q) Image is Erect
             principle focus of a converging lens
    (C)    Object is between optic center and 2nd         (r) Image is of greater size
             principle focus of a diverging lens                    than the object
    (D)     Object is between optic center and 2nd        (s) Image is of smaller size 
             principle focus of a converging lens                  than the object
                                                                              (t) Image distance is more than |f|

A double slit interference pattern is produced on a screen, as shown in the figure, using monochromatic light of wavelength 500 nm. Point P is the location of the central bright fringe, that is produced when light waves arrive in phase without any path difference. A choice of three strips A, B and C of transparent materials with different thicknesses and refractive indices is available, as shown in the table. These are placed over one or both of the slits, singularly or in conjunction, causing the interference pattern to be shifted across the screen from the original pattern. In the column-I, how the strips have been placed, is mentioned whereas in the column-II, order of the fringe at point P on the screen that will be produced due to the placement of the strip(s), is shown. Correctly match both the columns.

The compound 'X' is :

Observe the following reaction and identify the end product.

The major product of following sequence of reactions is

The final product of the following reaction is