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P Pankaj Sanodiya
Given Virtual image area = 6.25 mm2 Actual ara = 1 mm2 We can calculate linear magnification as  we also know  Now, according to the lens formula  and  Since the image is forming at a distance which is less than 25 cm, it can not be seen by eye distinctively.

P Pankaj Sanodiya
a) maximum magnifying is possible when our image distance will be equal to minimum vision point that is,      (Given) Now according to the lens formula Hence required object distance for viewing squares distinctly is 7.14 cm away from the lens. b) Magnification of the lens:   c) Magnifying power    Since the image is forming at near point ( d = 25 cm ), both magnifying power and...

P Pankaj Sanodiya
Given The focal length of the convex lens  here liquid is acting like the mirror so, the focal length of the liquid  the focal length of the system(convex + liquid)  Equivalent focal length when two optical systems are in contact   Now, let us assume refractive index of the lens be  The radius of curvature are  and . As we know, Now, let refractive index of liquid be  The radius of...

P Pankaj Sanodiya
Given Angle of deflection  The distance of the screen from the mirror  The reflected rays will bet deflected by twice angle of deviation that is   Now from the figure, it can be seen that Hence displacement of the reflected spot of the light is .

P Pankaj Sanodiya
Given, Distance between the objective mirror and secondary mirror  The radius of curvature of the Objective Mirror    So the focal length of the objective mirror   The radius of curvature of the secondary mirror    so, the focal length of the secondary mirror   The image of an object which is placed at infinity, in the objective mirror, will behave like a virtual object for the secondary...

P Pankaj Sanodiya
Given, image is formed at a distance  = 25cm As we know, magnification of eyepiece lens is given by : Now,  Height of the final image is given by : Therefore, the height of the final image will be 28.2 cm

P Pankaj Sanodiya
Given, focal length of the objectove lens = = 140cm focal length of the eyepiece lens =  = 5 cm Height of tower  = 100m Distance of object which is acting like a object  = 3km = 3000m. The angle subtended by the tower at the telescope  Now, let the height of the image of the tower by the objective lens is  . angle made by the image by the objective lens  : Since both, the angles are the same...

P Pankaj Sanodiya
a) Given, focal length of the objective lens = = 140cm focal length of the eyepiece lens =  = 5 cm  The separation between the objective lens and eyepiece lens is given by: Hence, under normal adjustment separation between two lenses of the telescope is 145 cm.

P Pankaj Sanodiya
Given, the focal length of the objective lens  the focal length of the eyepiece lens  normally, least distance of vision = 25cm Now, as we know magnifying power when the image is at d = 25 cm is Hence magnification, in this case, is 33.6.

P Pankaj Sanodiya
Given, the focal length of the objective lens  the focal length of the eyepiece lens  normally, least distance of vision = 25cm Now, As we know magnifying power:   Hence magnifying power is 28.

P Pankaj Sanodiya
Given, magnifying power = 30 objective lens focal length = 1.25cm eyepiece lens focal length = 5 cm Normally, image is formed at distance d = 25cm Now, by the formula; Angular magnification by eyepiece:    From here, magnification by the objective lens :      since   ( ) According to the lens formula: from here, hence object must be 1.5 cm away from the objective lens. Now for the...

P Pankaj Sanodiya
When we view through a compound microscope, our eyes should be positioned a short distance away from the eyepiece lens for seeing a clearer image. The image of the objective lens in the eyepiece lens is the position for best viewing. It is also called "eye-ring" and all reflected rays from lens pass through it which makes it the ideal position for the eye for the best view. When we put our eyes...

P Pankaj Sanodiya
We need more magnifying power and angular magnifying power in a microscope in order to use it effectively. Keeping both objective focal length and eyepiece focal length small makes the magnifying power greater and more effective.

P Pankaj Sanodiya
Firstly, grinding a lens with very small focal length is not easy and secondly and more importantly, when we reduce the focal length of a lens, spherical and chronic aberration becomes more noticeable. they both are defects of the image, resulting from the ways of rays of light.

P Pankaj Sanodiya
Yes, angular magnification will change if we move our eye away from the lens. this is because then angle subtended by lens would be different than the angle subtended by eye. When we move our eye form lens, angular magnification decreases. Also, one more important point here is that object distance does not have any effect on angular magnification.

P Pankaj Sanodiya
Angular magnification is the ratio of tangents of the angle formed by object and image from the centre point of the lens. In this question angle formed by the object and a virtual image is same but it provides magnification in a way that, whenever we have object place before 25cm, the lens magnifies it and make it in the vision range. By using magnification we can put the object closer to the...

P Pankaj Sanodiya
Given, Object distance u = -9cm Focal length of convex lens = 10cm According to the lens formula a) Magnification The area of each square in the virtual image b) Magnifying power c) No,  . Both the quantities will be equal only when image is located at the near point |v| = 25 cm

P Pankaj Sanodiya
Given  Object height = 1.5 cm  Object distance from convex lens = -40cm According to lens formula Magnificatio due to convex lens:   The image of convex lens will act as an object for concave lens, so, Magnification due to concave lens : The combined magnification:   Hence height of the image =   = 0.652 * 1.5 = 0.98cm Hence height of image is 0.98cm.