- A point object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices $n_1$ and $n_2(n_2 > n_1)$. Draw the ray diagram and deduce the relation between the object distance (u), image distance (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium.
- A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, find its new focal length.
CBSE FOREIGN - SET 2 2017
Download our app for free and get started
For small angles
$\tan\angle NOM=\frac{MN}{OM} :\tan\angle NCM=\frac{MN}{NC}$
and $\tan\angle NIM=\frac{MN}{MI}$
For Δ???, i is exterior angle, therefore
$\text{i}=\angle NOM+\angle NCM=\frac{MN}{OM}+\frac{MN}{MC}$
Similarly $\text{r}=\frac{MN}{MC}-\frac{MN}{MI}$
For small angles Snells law can be written as
$n_1\text{i}=n_2r$
$\therefore\text{ }\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}$
$\therefore\text{ }\text{OM}=\text{-u},\text{ }\text{MI}=+\text{v},\text{ }\text{MC}=+\text{R}$ (using sign conversion)
$\therefore\text{ }\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$
- Lens Maker’s formula is
$\therefore\frac{1}{20}=(1.6-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\therefore\big(\frac{1}{R_1}-\frac{1}{R_2}\big)=\frac{1}{20\times0.6}=\frac{1}{12}$
Let f be the focal length of the lens in water
$\therefore\frac{1}{f'}=\frac{1.6-1.3}{1.3}\big(\frac{1}{R_1}-\frac{1}{R_2}\big)=\frac{0.3}{12\times1.3}$
$\frac{1}{f'}=\frac{120\times1.3}{3}=52\text{ }cm$
Download our appand 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.*
Similar Questions
- 1Consider 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
- 2View SolutionA compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece 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), and (b) at infinity? What is the magnifying power of the microscope in each case?
- 3View SolutionMrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new spects, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker.
- Write two qualities displayed each by Anuja and her mother.
- How do you explain this fact using lens maker's formula?
- 4View SolutionA compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5cm and the tube length is 6.5cm. Find the focal length of the eyepiece.
- 5View SolutionA normal eye has retina 2cm behind the eye-lens. What is the power of the eye-lens when the eye is
- Fully relaxed,
- Most strained?
- 6A point object 'O' is kept in a medium of refractive index $n_1$ in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index $n_2$ from the first one, as shown in the figure.View Solution
- Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of $n_1, n_2$ and R.
- When the image formed above acts as a virtual object for concave spherical, surface separating the medium $n_2$ from $n_1 (n_2> n_1)$, draw this ray diagram and write the similar (similar to (a) relation. Hence obtain the expression for the lens maker's formula.
- 7View SolutionDefine magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece. - 8View SolutionA pin of length 2.0cm lies along the principal axis of a converging lens, the centre being at a distance of 11cm from the lens. The focal length of the lens is 6cm. Find the size of the image.
- 9View SolutionUse the mirror equation to deduce that:
- an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
- a convex mirror always produces a virtual image independent of the location of the object.
- the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
- 10View Solution
- For a ray of light travelling from a denser medium of refractive index $n_1$ to a rarer medium of refractive index $n_2$ prove that $\frac{\text{n}_{2}}{\text{n}_{1}} = \sin \text{i}_{c} ,$ where $i_c$ is the critical angle of incidence for the media.
- Explain with the help of a diagram, how the above principle is used for transmission of video signals using optical fibres.