Screen should be placed at 54 cm from the mirror on the same side as the object.
Minus sing indicates that image is inverted.
When the candle is moved closer to mirror, the screen has to be moved away from the mirror. However, when candle is at a distance less then 18 cm from the mirror, image formed would be virtual and screen is not required, as the virtual image cannot be taken on the screen.
Image is virtual, formed at 6.7 cm at the back of the mirror.
Image is erect, and of course virtual.
As needle is moved farther from the mirror, image moves away from the mirror ( upto F) and goes on decreasing in size.
Distance through which microscope has to moved up = 9.4 – 7.67 =1.73 cm</div
Sol. In the source of light (S) is 80 cm below the surface of water i.e., SO = 80 cm = 0.8 m.
Area of the surface of water through which light from the bulb can emerge is area of the circle of radius
Q.8. A beam of light converges to a point P.A lens is placed in the path of the convergent beam 12 cm form P. At what point does the beam converge if the lens is(a) a convex lens of focal length 20 cm. (b) a concave lens of focal length 16 cm ?
Sol. Here, the point P on the right of the lens acts as a virtual object,
Hence, image is at 48 cm. to the right of the lens, where the beam would converge.
As the object is moved away from the lens, virtual image moves towards focus of lens (but never beyond focus). The size of image goes on decreasing.
F= – 60 cm
The combination of lenses behaves as a concave les. The system is not converging.
Q.11. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece of focal length 6.25 cm, separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm) (b) at infinity ?
What is the magnifying power of the microscope in each case ?
Q.14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of telescope ?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 ⨯ 106 m and radius of lunar orbit is 3.8 ⨯ 108 m.
Q.15. Use the mirror equation to deduce that :
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2 f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
The answer does not depend upon the location of the slab.
Q.17. (a) shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of incident rays with the axis of the pipe for which total reflections inside the pipe take place as shown.
(b) What is the answer if there is no outer covering of the pipe ?
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we see a virtual image, we bring it to the screen i.e. retina of our eye. Is there a contradiction ?
(c) A diver under water looks obliquely at a fisherman standing om the bank of a lake. Would the fisherman loo taller or shorter then what he actually is ?
(d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ?
(e) The refractive index of diamond is much greater then that of ordinary glass. Is this fact of some use to a diamond cutter ?
(b) No, there is no contradiction. Eye lens is convergent. If form a real image of the virtual object (i.e., the virtual image being seen) on the retina.
(c) As fisherman is in air, light travels from rare to tenser medium. It bends towards the normal, appearing to come from a larger distance. Therefore, to the diver under water, fisherman looks taller.
(d) Yes, the apparent depth decreases further, when water tank is viewed obliquely compared to the depth when seen near normally.
As refractive index of diamond is much greater then that of ordinary glass, critical angle C for diamond is much smaller ( ) as compared to that for glass ( ). A skilled diamond cutter exploits the large range of angles of incidence of light (240 to 900) to ensure that light entering the diamond suffers multiple total internal reflections within the diamond. This produces sparkling effect in the diamond.
(b) As object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and convex lens is 40 cm. Determine the magnification produced by the two lens system and size of image.
(i) deviate a pencil of white light without much dispersion.
(ii) disperse and displace a pencil of white light without much deviation.
Taking crown glass prism of certain angle, we go on increasing angle of flint glass prims till this condition is met. In the final combination however, angle of flint glass prism will be smaller than the angle of crown glass prism as for flint glass is more then for crown glass.
Distance between cornea/eyes lens and retina = focal length of eye
Hence, range of accommodation of eye lens is roughly 20 to 24 dioptre.
However, when eye ball has normal length, but the eye lens loses partially its power of accommodation, the defect is called presbyopia.
The objects at infinity produce virtual images at 100 cm (using spectacles). To see objects between 25 cm and 100 cm, the person uses the ability of accommodation of his eye lens. This ability is partially lost in old age. The near point of the eye may recede to 50 cm. He has, therefore to use glasses of suitable power for reading.
to ? How is such a defect of vision corrected ?
This defect is removed by using a cylindrical lens with its axis along the vertical.
(a) What are the closest and farthest distances at which he can read the book, when viewing through the magnifying glass ?(b) What is the maximum and the minimum angular magnifications (magnifying powers) possible using the above simple microscope ?
For the closest distance ; v = -25 cm
(a) What is the magnification (magnifying power) of the lens ?
(b) Is the magnification in (a) equal to the magnifying power in (b) ? Explain
(c) No, magnification in (a) which is (v/u) cannot be equal to magnifying power in (b) which is (d/u)
Unless v = d i.e., image is located at the least distance of distance vision.
(ii) What is the magnification in this case ?
(iii) Is the magnification equal to magnifying power in this cas ? Explain.
Yes, the magnification and magnifying power in this case are equal, because image is formed at the least distance of distinct vision.
As the virtual image is at 15 cm ; whereas distance of distinct vision is 25 cm, therefore, the image cannot be seen distinctly by the eye.
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then dose a magnifying glass provide angular magnification ?
(b) In viewing through a magnifying glass, one usually positions one’s eye very close to the lens.
Does angular magnification change if the eye is moved back ?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power ?
(d) Why must both the objective and eye piece of a compound microscope have short focal lengths ?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye piece, but a short distance away from it for best viewing, why ? How much should be that short distance between the eye and eye piece ?
By using magnifying glass, we keep the object far more closer to the eye than at 25 cm, its normal position without use of glass. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved.
(b) Yes, the angular magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less then the angle subtended at the lens. The effect is negligible when image is at much larger distance.
(c) Theoretically, it is true. However, when we decrease focal length, aberration both spherical and chromatic become more pronounced. Further, is it difficult to grind lenses of very small focal lengths.
(e) The image of objective lens in eye piece is called ‘eye ring’. All the rays from the object refracted by the objective go through the eye ring. Therefore, ideal position for our eyers for viewing is this eye ring only.
When eye is too close to the eye piece, field of view reduced and eyes do not collect much of the light. The precise location of the eye ring would depend upon the separation between the objective and eye piece, and also on focal length of the eye piece.
(a) the telescope is in normal adjustment (i.e. when the final image is at infinity)
(b) the final image is formed at the least distance of distinct vision (25 cm).
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ?
(c) What is the height of the final image of the tower if it is formed at 25 cm ?
Radius of curvature of secondary mirror
R2 = 140 mm ;
along the axis until its inverted image is found at the position
of the needle. The distance of the needle from the lens is
measured to be 45.0 cm. The liquid is removed and the
experiment is repeated. The new distance is measured to
be 30.0 cm. What is the refractive index of the liquid ?
Combined focal length, F = 45. 0 cm