Question
A simple microscope is rated 5 for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40cm?

Answer

The simple microscope has magnification of 5 for normal relaxed eye (D = 25cm).Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$
For the relaxed farsighted eye, $\text{D} = 40\text{cm}$
So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$
So, its magnifying power is 8X.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if:
  1. The velocity is perpendicular to the diameter joining free ends.
  2. The velocity is parallel to this diameter.
Electric potential at a point $P , V (x, y, z)=$ $6 x-8 x y^2-8 y+6 y z-4 z^2$. Find the electric force on a 4 coulomb point charge located at the origin.
When two surfaces are polished, the friction coefficient between them decreases. But the friction coefficient increases and becomes very large if the surfaces are made highly smooth. Explain.
An object is placed at the principal focus of a concave lens of focal length f. Where will its image be formed?
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
Is it true that the reaction of a gravitational force is always gravitational, of an electromagnetic force is always electromagnetic and so on?
A convex lens made of a material of refractive index $n_1$ is kept in a medium of refractive index $n_2$. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if : $(i)\  n_1 > n_2 \ (ii)\  n_1 = n_2 \ (iii)\  n_1< n_2.
\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}\Big)$
Let $\vec{\text{a}}=4\vec{\text{i}}+3\vec{\text{j}}$ and $\vec{\text{b}}=3\vec{\text{i}}+4\vec{\text{j}}.$ (a) Find the magnitudes of:
  1. $\vec{\text{a}}$
  2. $\vec{\text{b}}$
  3. $\vec{\text{a}}+\vec{\text{b}}$
  4. $\vec{\text{a}}-\vec{\text{b}}$
A conducting disc of radius r rotates with a small but constant angular velocity $\omega$ about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the centre and the periphery of the disc.
The torque of a force $\overrightarrow{\text{F}}$ about a point is defined as $\overrightarrow{\text{r}}=\overrightarrow{\text{r}}\times\overrightarrow{\text{F}}.$ Suppose $\overrightarrow{\text{r}}, \overrightarrow{\text{F}}$and $\overrightarrow{\text{r}}$ are all nonzero. Is $\text{r}\times\overrightarrow{\text{r}}\Bigg|\Bigg|\overrightarrow{\text{F}}$ always true? Is it ever true?
What is being done to change the range of voltmeter?