Advertisements
Advertisements
Question
A double convex lens of + 5 D is made of glass of refractive index 1.55 with both faces of equal radii of curvature. Find the value of its radius of curvature.
Advertisements
Solution
From lens maker formula, we have
\[P = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]
\[\text { where} \]
\[ P =\text { Power of lens } = + 5 D\]
\[\mu = \text { Refractive index of the lens } = 1 . 55\]
\[ R_1 =\text { Radius of curvature of first face } \left( + ve \right)\]
\[ R_2 = \text { Radius of curvature of second face } \left( - ve \right)\]
\[\text {Given }: \]
\[ R_1 = R_2 = R\]
\[ \Rightarrow P = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]
\[ \Rightarrow 5 = (1 . 55 - 1)\left( \frac{1}{R} - \frac{1}{- R} \right)\]
\[ \Rightarrow 5 = (1 . 55 - 1)\left( \frac{2}{R} \right)\]
\[ \Rightarrow 5 = 0 . 55\left( \frac{2}{R} \right)\]
\[ \Rightarrow R = \frac{0 . 55 \times 2}{5}\]
\[ \Rightarrow R = 0 . 22 m\]
APPEARS IN
RELATED QUESTIONS
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
- Determine the ‘effective focal length’ of the combination of the two lenses, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of the effective focal length of this system useful at all?
- An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system and the size of the image.
An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image
A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
- What is the magnification produced by the lens? How much is the area of each square in the virtual image?
- What is the angular magnification (magnifying power) of the lens?
- Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved 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?
Two concave lenses L1 and L2 are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination
Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without loosing the energy of the light. This increase the intensity. Describe how the converging lenses should be placed to do this.
A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sum is just setting? Refractive index of water = 4/3.
In the given figure the radius of curvature of the curved face in the planoconvex and the planoconcave lens is 15 cm each. The refractive index of the material of the lenses is 1.5. Find the final position of the image formed.
