A card sheet divided into squares each of size $1\ mm^2$ 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.
Exercise
Download our app for free and get started
- Given,
Focal length of converging lens, f = 10 cm
Now, Using the formula,
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$\Rightarrow \ \frac{1}{\text{v}}=\frac{1}{10}+\frac{1}{-9}$
$=\frac{-9+10}{-90}$
$=\frac{1}{-90}$
i.e., v = -90 cm
Magnification, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{-90}{-9}=10$
Therefore,
Area of the each square in the virtual image $= (10 \times 10 \times 1) mm^2 = 100 mm^2 = 1 cm^2.$
- Given,
Object distance, u = -9 cm
Therefore,
Angular magnification $=\frac{\text{D}}{\text{u}}=\frac{-25}{-9}=2.78$
- No. Magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are different from each other.
Thus, magnification magnitude $=|\frac{\text{v}}{\text{u}}|$ and,
Magnifying power is $\frac{25}{|\text{u}|}=\frac{\text{D}}{\text{u}}$.
Only when the image is located at the least distance of distinct vision, we can have both the quantities equal.
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
- 1View SolutionA professor reads a greeting card received on his 50th birthday with +2.5D glasses keeping the card 25cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50cm away. What power of lens should he now use?
- 2View SolutionAt what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
- 3View SolutionAn eye can distinguish between two points of an object if they are separated by more than 0.22mm when the object is placed at 25cm from the eye. The object is now seen by a compound microscope having a 20D objective and 10D eyepiece separated by a distance of 20cm. The final image is formed at 25cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?
- 4k transparent slabs are arranged one over another. The refractive indices of the slabs are $\mu_1,\mu_2,\mu_3, \ ...\mu_{\text{k}}$ and the thicknesses are $t_1, t_2, t_3, ... t_k$ . An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.View Solution
- 5Show that for a material with refractive index $\mu\geq\sqrt{2}$, light incident at any angle shall be guided along a length perpendicular to the incident face.View Solution
- 6View 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.
- 7If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refrative index of the medium given byView Solution
$\text{n}(\text{r})=1+2\frac{\text{GM}}{\text{rc}^2}$
where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object. - 8View SolutionOne day Chetan’s mother developed a severe stomach ache all of a sudden. She was rushed to the doctor who suggested for an immediate endoscopy test and gave an estimate of expenditure for the same. Chetan immediately contacted his class teacher and shared the information with her. The class teacher arranged for the money and rushed to the hospital. On realising that Chetan belonged to a below average income group family, even the doctor offered concession for the test fee. The test was conducted successfully.
Answer the following questions based on the above information:- Which principle in optics is made use of in endoscopy?
- Briefly explain the values reflected in the action taken by the teacher.
- In what way do you appreciate the response of the doctor on the given situation?
- 9Consider the situation shown in figure. The elevator is going up with an acceleration of $2.00m/s^2$ and the focal length of the mirror is 12.0cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t = 0 when the distance of B from the mirror is 42.0cm. Find the distance between the image of the block B and the mirror at $t = 0.200s$. Take $g = 10m/s^2$.View Solution
- 10A 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.