मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

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 - Physics

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

पर्याय

  • 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.

MCQ

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:
\[x = \frac{(2f + 40) \pm \sqrt{(2f + 40 )^2 - 160f}}{2}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 18: Geometrical Optics - MCQ [पृष्ठ ४१२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 18 Geometrical Optics
MCQ | Q 7 | पृष्ठ ४१२

संबंधित प्रश्‍न

What does sign of power (+ve or –ve) indicate?


Define 1 dioptre of power of a lens.


A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?


Give the usual name for the following:
A point inside a lens through which the light passes undeviated.


Which of the two has a greater power: a lens of short focal length or a lens of large focal length?


The focal length of a convex lens is 25 cm. What is its power? 


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?


The focal length of a lens is +150 mm. What kind of lens is it and what is its power? 


The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm :
(a) calculate the focal length of lens Y.
(b) state the nature of lens Y.

 


A diverging lens has a focal length of 0.10 m. The power of this lens will be:


Define the term power of a lens.


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 convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image?


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?


Express nanometer (nm) in terms of Angstrom

The power of a lens is +2.0 D. Find its focal length and state the kind of lens.


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.


Complete the diagram to show the formation of the image of the object AB.

(i) Name the Lens LL’.
(ii) Where is the image of the object AB formed?
(iii) State three characteristics of the image.


What is power of accommodation of eye?


The above lens has a focal length of 10 cm. The object of height 2 mm is placed at a distance of 5 cm from the pole. Find the height of the image.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×