Advertisements
Advertisements
प्रश्न
A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and its is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
विकल्प
must be less than 10 cm
must be greater than 20 cm
must not be greater than 20 cm
must not be less than 10 cm.
Advertisements
उत्तर
must be greater than 20 cm
Let the image be formed at a distance of x cm from the lens.
Therefore, the distance of the object from the lens, u, will be = (40 − x) cm
From lens formula:
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
\[ \Rightarrow \frac{1}{f} = \frac{1}{x} - \frac{1}{40 - x}\]
\[ \Rightarrow \frac{1}{f} = \frac{40 - x - x}{40x - x^2}\]
\[ \Rightarrow f = \frac{40x - x^2}{40 - 2x}\]
\[ \Rightarrow f(40 - 2x) = 40x - x^2 \]
\[ \Rightarrow x^2 - 2fx - 40x + 40f = 0\]
\[ \Rightarrow x^2 - (2f + 40)x + 40f = 0\]
Therefore, we get x as:
\[x = \frac{(2f + 40) \pm \sqrt{(2f + 40 )^2 - 160f}}{2}\]
APPEARS IN
संबंधित प्रश्न
What is meant by a power of a lens? Define its SI unit.
State power of a lens S.I. unit.
(a) At what distance should the lens be held from the figure in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.

Consider two statements A and B given below:
A: real image is always inverted
B: virtual image is always erect
Out of these two statements:
How is the power of a lens related to its focal length?
The optician's prescription for a spectacle lens is marked +0.5 D. What is the:
(a) nature of spectacle lens?
(b) focal length of spectacle lens?
Fill in the following blank with suitable word:
For converging lenses, the power is __________ while for diverging lenses, the power is ___________.
The power of a lens is + 1.0 D is :
Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5.
A normal eye is not able to see objects closer than 25 cm because
A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.
Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?
A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
Surabhi from std. X uses spectacle. The power of the lenses in her spectacle is 0.5 D.
Answer the following questions from the given information:
- Identify the type of lenses used in her spectacle.
- Identify the defect of vision Surabhi is suffering from.
- Find the focal length of the lenses used in her spectacle.
A lens forms an upright and diminished image of an object, irrespective of its position. What kind of lens is this? Draw an outline ray diagram to show the formation of the image. State the position and one more characteristic of the image.
The following diagram shows the object O and the image I formed by a lens. Copy the diagram and on it mark the positions of the lens LL’ and focus (F). Name the lens.

If the lens is of focal length 25 cm. Calculate the power of the lens.
Find the power of a convex lens of focal length of + 25 cm.
Assertion and reasoning type
- Assertion: Myopia is due to the increase in the converging power of eye lens.
- Reason: Myopia can be corrected with the help of concave lens.
