Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Physics-

General

Easy

Question

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point Which of the following statements are correct regarding spherical aberration :

It can be completely eliminated
If can’t be completely eliminated but it can’t be minimised by allowing either paraxial or marginal rays to hit the mirror
It is reduced by taking large aperture mirrors
none of the above

The correct answer is: If can’t be completely eliminated but it can’t be minimised by allowing either paraxial or marginal rays to hit the mirror

Related Questions to study

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point For paraxial rays, focal length approximately is

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point For paraxial rays, focal length approximately is

physics-General

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point The total deviation suffered by the ray falling on mirror at an angle of incidence equal to 60° is

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point The total deviation suffered by the ray falling on mirror at an angle of incidence equal to 60° is

physics-General

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point If f_p and f_m represent the focal length of paraxial and marginal rays respectively, then correct relationship is :

Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ‘f’ on angle of incidence ‘ q ’ as shown in figure is given by $f equals R minus fraction numerator K over denominator 2 end fraction s e c invisible function application theta$ where R is radius of curvature of mirror and q is the angle of incidence The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays As a result of above dependence different rays are brought to focus at different points and the image of a point object is on a point If f_p and f_m represent the focal length of paraxial and marginal rays respectively, then correct relationship is :

physics-General

parallel

Most materials have the refractive index, n > 1 So, when a light ray from air enters a naturally occurring material, then by Snell’s $text law, end text fraction numerator sin invisible function application theta subscript 1 end subscript over denominator sin invisible function application theta subscript 2 end subscript end fraction equals fraction numerator n subscript 1 end subscript over denominator n subscript 2 end subscript end fraction comma$ it is understood that the refracted ray bends towards the normal But it never emerges on the same side of the normal as the incident ray According to electromagnetism, the refractive index of the medium is given by the relation, $n equals open parentheses fraction numerator c over denominator V end fraction close parentheses equals plus-or-minus square root of epsilon subscript r end subscript mu subscript r end subscript end root$ , where c is the speed of the electromagnetic waves in vacuum, v its speed in the medium, e_r and m_r are negative, one must choose the negative root of n Such negative refractive index materials can now be artificially prepared and are called metamaterials They exhibit signficantly different optical behaviour, without violating any physical laws Since n is negative, it results in a change in the direction of propagation of the refracted light However, similar to normal materials, the frequency of light remains unchanged upon refraction even in metamaterials For light incident from air on a meta-material, the appropriate ray diagrams

Most materials have the refractive index, n > 1 So, when a light ray from air enters a naturally occurring material, then by Snell’s $text law, end text fraction numerator sin invisible function application theta subscript 1 end subscript over denominator sin invisible function application theta subscript 2 end subscript end fraction equals fraction numerator n subscript 1 end subscript over denominator n subscript 2 end subscript end fraction comma$ it is understood that the refracted ray bends towards the normal But it never emerges on the same side of the normal as the incident ray According to electromagnetism, the refractive index of the medium is given by the relation, $n equals open parentheses fraction numerator c over denominator V end fraction close parentheses equals plus-or-minus square root of epsilon subscript r end subscript mu subscript r end subscript end root$ , where c is the speed of the electromagnetic waves in vacuum, v its speed in the medium, e_r and m_r are negative, one must choose the negative root of n Such negative refractive index materials can now be artificially prepared and are called metamaterials They exhibit signficantly different optical behaviour, without violating any physical laws Since n is negative, it results in a change in the direction of propagation of the refracted light However, similar to normal materials, the frequency of light remains unchanged upon refraction even in metamaterials For light incident from air on a meta-material, the appropriate ray diagrams

physics-General

The element in the first row and third column of the inverse of the matrix $open square brackets table row 1 2 cell negative 3 end cell row 0 1 2 row 0 0 1 end table close square brackets$ is

The element in the first row and third column of the inverse of the matrix $open square brackets table row 1 2 cell negative 3 end cell row 0 1 2 row 0 0 1 end table close square brackets$ is

If A and B are non-singular square matrices of same order, then $a d j left parenthesis A B right parenthesis$ is equal to

If A and B are non-singular square matrices of same order, then $a d j left parenthesis A B right parenthesis$ is equal to

parallel

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is

$C subscript 6 H subscript 5 minus C H equals C H C H O not stretchy rightwards arrow with X on top C subscript 6 H subscript 5 C H equals C H C H subscript 2 O H$ In the above sequence X can be

$C subscript 6 H subscript 5 minus C H equals C H C H O not stretchy rightwards arrow with X on top C subscript 6 H subscript 5 C H equals C H C H subscript 2 O H$ In the above sequence X can be

chemistry-General

A ray of light traveling in air is incident at grazing angle (Ð >i 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m The point of incidence is the medium A (0, 0) The medium has a variable index of refraction n(y) given by $n left parenthesis y right parenthesis equals open square brackets k y to the power of 3 divided by 2 end exponent plus 1 close square brackets to the power of 1 divided by 2 end exponent$ where $k equals 1.0 left parenthesis m right parenthesis to the power of negative 3 divided by 2 end exponent$ The refractive index of air is 1 The coordinates (x1, y(A) of the point P where the ray intersects the upper surface of the slabair boundary are

A ray of light traveling in air is incident at grazing angle (Ð >i 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m The point of incidence is the medium A (0, 0) The medium has a variable index of refraction n(y) given by $n left parenthesis y right parenthesis equals open square brackets k y to the power of 3 divided by 2 end exponent plus 1 close square brackets to the power of 1 divided by 2 end exponent$ where $k equals 1.0 left parenthesis m right parenthesis to the power of negative 3 divided by 2 end exponent$ The refractive index of air is 1 The coordinates (x1, y(A) of the point P where the ray intersects the upper surface of the slabair boundary are

physics-General

parallel

A ray of light traveling in air is incident at grazing angle (Ð >i 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m The point of incidence is the medium A (0, 0) The medium has a variable index of refraction n(y) given by $n left parenthesis y right parenthesis equals open square brackets k y to the power of 3 divided by 2 end exponent plus 1 close square brackets to the power of 1 divided by 2 end exponent$ where $k equals 1.0 left parenthesis m right parenthesis to the power of negative 3 divided by 2 end exponent$ The refractive index of air is 1 Equation for the trajectory y(x) of the ray in the medium is

A ray of light traveling in air is incident at grazing angle (Ð >i 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m The point of incidence is the medium A (0, 0) The medium has a variable index of refraction n(y) given by $n left parenthesis y right parenthesis equals open square brackets k y to the power of 3 divided by 2 end exponent plus 1 close square brackets to the power of 1 divided by 2 end exponent$ where $k equals 1.0 left parenthesis m right parenthesis to the power of negative 3 divided by 2 end exponent$ The refractive index of air is 1 Equation for the trajectory y(x) of the ray in the medium is

physics-General

A ray of light traveling in air is incident at grazing angle (Ð >i 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m The point of incidence is the medium A (0, 0) The medium has a variable index of refraction n(y) given by $n left parenthesis y right parenthesis equals open square brackets k y to the power of 3 divided by 2 end exponent plus 1 close square brackets to the power of 1 divided by 2 end exponent$ where $k equals 1.0 left parenthesis m right parenthesis to the power of negative 3 divided by 2 end exponent$ The refractive index of air is 1 The incident angle at B(x, y) in the medium and the slope at B are related by the formula

A ray of light traveling in air is incident at grazing angle (Ð >i 90º) on a long rectangular slab of a transparent medium of thickness t = 1.0 m The point of incidence is the medium A (0, 0) The medium has a variable index of refraction n(y) given by $n left parenthesis y right parenthesis equals open square brackets k y to the power of 3 divided by 2 end exponent plus 1 close square brackets to the power of 1 divided by 2 end exponent$ where $k equals 1.0 left parenthesis m right parenthesis to the power of negative 3 divided by 2 end exponent$ The refractive index of air is 1 The incident angle at B(x, y) in the medium and the slope at B are related by the formula

physics-General

A point object ‘O’ is placed along x-axis An equi-convex thin lens $open parentheses mu subscript g end subscript equals 1.5 close parentheses$ of focal length f=20cm in air is placed so that its principal axis is along x-axis Now the lens is cut at the middle (along the principal axis) and upper half is shifted along x-axis and y-axis by 20cm and 2mm respectively and right side of lower half is filled with water $open parentheses mu subscript w end subscript equals fraction numerator 4 over denominator 3 end fraction close parentheses$ co-ordinates of image formed by lenses are

A point object ‘O’ is placed along x-axis An equi-convex thin lens $open parentheses mu subscript g end subscript equals 1.5 close parentheses$ of focal length f=20cm in air is placed so that its principal axis is along x-axis Now the lens is cut at the middle (along the principal axis) and upper half is shifted along x-axis and y-axis by 20cm and 2mm respectively and right side of lower half is filled with water $open parentheses mu subscript w end subscript equals fraction numerator 4 over denominator 3 end fraction close parentheses$ co-ordinates of image formed by lenses are