Derive the lens formula, $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ for a concave lens, using the necessary ray diagram. Two lenses of powers 10 D and – 5 D are placed in contact. 1. Calculate the power of the new lens. 2. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Diagram: Derivation: $\frac{1}{\text{v}} -\frac{1}{\text{u}} = \frac{1}{\text{f}}$ 1. $\text{Power} = \text{P}_{1} + \text{P}_{2}$ $ = 10 - 5 = 5\text{D}$ 2. f = 20 cm $\text{m} = \frac{\text{v}}{\text{u}} = 2$ $\frac{1}{\text{f}} =\frac{1}{\text{v}} - \frac{1}{\text{u}}$ $\text{u} = -10\text{cm}.$

Derive the lens formula, $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ for a concave lens, using the necessary ray diagram. Two lenses of powers 10 D and – 5 D are placed in contact. 1. Calculate the power of the new lens. 2. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Diagram: Derivation: $\frac{1}{\text{v}} -\frac{1}{\text{u}} = \frac{1}{\text{f}}$ 1. $\text{Power} = \text{P}_{1} + \text{P}_{2}$ $ = 10 - 5 = 5\text{D}$ 2. f = 20 cm $\text{m} = \frac{\text{v}}{\text{u}} = 2$ $\frac{1}{\text{f}} =\frac{1}{\text{v}} - \frac{1}{\text{u}}$ $\text{u} = -10\text{cm}.$

Derive the lens formula, $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ for a concave lens, using the necessary ray diagram. Two lenses of powers 10 D and – 5 D are placed in contact. 1. Calculate the power of the new lens. 2. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Diagram: Derivation: $\frac{1}{\text{v}} -\frac{1}{\text{u}} = \frac{1}{\text{f}}$ 1. $\text{Power} = \text{P}_{1} + \text{P}_{2}$ $ = 10 - 5 = 5\text{D}$ 2. f = 20 cm $\text{m} = \frac{\text{v}}{\text{u}} = 2$ $\frac{1}{\text{f}} =\frac{1}{\text{v}} - \frac{1}{\text{u}}$ $\text{u} = -10\text{cm}.$

Derive the lens formula, frac{1}{text{f}} = frac{1}{text{v}} - frac{1}{text{u}} for a concave lens, using the necessary ray diagram. Two lenses of powers 10 D and

5 Marks Questions

GSEB - English Medium

Derive the lens formula, $\frac{1}{\text{f}} = \frac{1}{\text{v}} - \frac{1}{\text{u}}$ for a concave lens, using the necessary ray diagram.
Two lenses of powers 10 D and – 5 D are placed in contact.

Calculate the power of the new lens.
Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

CBSE OUTSIDE DELHI - SET 1 2008

Download our app for free and get started

Solution

Diagram:

Derivation: $\frac{1}{\text{v}} -\frac{1}{\text{u}} = \frac{1}{\text{f}}$

$\text{Power} = \text{P}_{1} + \text{P}_{2}$

$ = 10 - 5 = 5\text{D}$

f = 20 cm

$\text{m} = \frac{\text{v}}{\text{u}} = 2$

$\frac{1}{\text{f}} =\frac{1}{\text{v}} - \frac{1}{\text{u}}$

$\text{u} = -10\text{cm}.$

STD 12 Science
Physics
Ray Optics and Optical Instruments
NCERT

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Download App

Similar Questions

1
An object is placed at a distance of 30cm from a converging lens of focal length 15cm. A normal eye (near point 25cm, far point infinity) is placed close to the lens on the other side.
1. Can the eye see the object clearly?
2. What should be the minimum separation between the lens and the eye so that the eye can clearly see the object?
3. Can a diverging lens, placed in contact with the converging lens, help in seeing the object clearly when the eye is close to the lens?
View Solution
2
Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are $\mu_\text{r},\mu_\text{y}$ and $\mu_\text{v}$ respectively and those for the flint glass are $\mu'_\text{r},\mu'_\text{y}$ and $\mu'_\text{v}$ respectively. Find the ratio $\frac{\text{A}'}{\text{A}}$ for which.
1. There is no net angular dispersion.
2. There is no net deviation in the yellow ray.
View Solution
3
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?
View Solution
4
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
View Solution
5
1. Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope.
2. An astronomical telescope has an objective lens of focal length 20m and eyepiece of focal length 1cm.
1. Find the angular magnification of the telescope.
2. If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is $3.5 \times 10^6m$ and radius of lunar orbit is $3.8 \times 10^8m.$
View Solution
6
A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index $\mu.$ The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for $\text{t}<\sqrt{\frac{2\text{h}}{\text{g}}}.$ Consider only the image by a single refraction.
View Solution
7
A mass m = 50g is dropped on a vertical spring of spring constant 500N/m from a height h = 10cm as shown in figure. The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12cm facing the mass is fixed with its principal axis coinciding with the line of motion of the mass, its pole being at a distance of 30cm from the free end of the spring. Find the length in which the image of the mass oscillates.

View Solution
8
Consider a thin lens placed between a source (S) and an observer (O) (Fig). Let the thickness of the lens vary as $\text{w}\text{(b)}=\text{w}_0-\frac{\text{b}^2}{\alpha}$, where b is the verticle distance from the pole. $w_0$ is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

View Solution
9
A diverging lens of focal length 20cm and a converging lens of focal length 30cm are placed 15cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
View Solution
10
A cylindrical vessel, whose diameter and height both are equal to 30cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible. The particle P is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible?

View Solution

Download our appand get started for free

Similar Questions

Download our app
and get started for free