Q: Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

\$\textbf{Explanation:}\$ \$\bullet\$A basic microscope can also be made with a convex lens. \$\bullet\$A virtual, erect, and magnified image is generated by keeping the object between the focal length and the mirror. \$\bullet\$A shorter focal length provides greater power and allows for higher magnification. \$\bullet\$As a result, a convex lens with a focal length of 5 cm is preferred.\$\textbf{Step C is correct.}\$

Q: If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is :

Using the lens formula for the objective lens,\$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \$\$ \dfrac{1}{1.2} = \dfrac{1}{v} - \dfrac{1}{-1.25} \$This gives \$ v = 30 cm\$Now, magnification is given \$m = \dfrac{v}{u} \times \dfrac{D}{{f}_{e}} \$Substituting the values, \$m = \dfrac{30}{1.25} \times \dfrac{25}{3} \$Thus, \$m = 200\$

Q: A simple microscope has a focal length of 5cm. The magnification at the least distance of distinct vision is

For single microscope,\$m= 1+\dfrac{D}{f}\$ \$= 1+\dfrac{25}{5}\$ \$= 6\$

Q: The focal length of a convex lens is 10 cm. The magnifying powers when it is used as a magnifying glass to form the image at (i) near point and (ii) far point are :

For the near point, magnification, \$ {m}_{1} = 1 + \dfrac{D}{f} \$For the far point, \$ {m}_{2} = \dfrac{D}{f} \$Thus, putting \$D = 25\$ and \$f = 10\$We get, \$ {m}_{1} = 3.5\$ and \${m}_{2} =2.5 \$

Question 1

Draw a ray diagram of compound microscope, when final image is formed at the minimum distance of distinct vision.

Accepted Answer

It consist of two convex lenses,  one objective of very small focal length with short aperture. And one Eyepiece with moderate focal length and large aperture.

Question 2

Describe the construction of a compound microscope. Derive an expression for its total magnification. Draw a ray diagram for the formation of image by a compound microscope.

Accepted Answer

Construction of Compound microscope:A schematic diagram of a compound microscope is shown in Fig. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.For magnification, $m_0=\dfrac{h'}{h}=\dfrac{L}{f_0}$where we have used the result ,   $tan \beta=\dfrac{h}{f_0}=\dfrac{h'}{L}$Here $h&#8242;$ is the size of the first image, the object size being $h$ and fo being the focal length of the objective. The first image is formed near the focal point of the eyepiece. As the first inverted image is near the focal point of the eyepiece.$m_e=1+\dfrac{D}{f_e}$When the final image is formed at infinity, the angular magnification due to the eyepiece.$m_e=\dfrac{D}{f_e}$Thus, the total magnification  when the image is formed at infinity, is$m=m_0m_e=\dfrac{L}{f_0} \dfrac{D}{f_e}$

Question 3

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

Accepted Answer

$\textbf{Explanation:}$ $\bullet$A basic microscope can also be made with a convex lens. $\bullet$A virtual, erect, and magnified image is generated by keeping the object between the focal length and the mirror. $\bullet$A shorter focal length provides greater power and allows for higher magnification. $\bullet$As a result, a convex lens with a focal length of 5 cm is preferred.$\textbf{Step C is correct.}$

Question 4

Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain.

Accepted Answer

Magnifying power is defined as the angle subtends at the eye by the image to the angle subtended by the object.$m=m_0\times m_e=\dfrac{L}{f_0}\times \dfrac{D}{f_e}$To increase the magnifying power both objective and eye piece must have short focal lengths such as,$m=\dfrac{L}{f_0}\times \dfrac{D}{f_e}$

Question 5

If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is :

Accepted Answer

Using the lens formula for the objective lens,$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $$ \dfrac{1}{1.2} = \dfrac{1}{v} - \dfrac{1}{-1.25} $This gives  $ v = 30 cm$Now, magnification is given $m =  \dfrac{v}{u} \times \dfrac{D}{{f}_{e}} $Substituting the values, $m = \dfrac{30}{1.25} \times \dfrac{25}{3} $Thus, $m = 200$

Question 6

Draw a labelled diagram of an image formed by a compound microscope, with the image at least distance of distinct vision. Write any one expression for its magnifying power.

Accepted Answer

Expression of magnifying power of a compound microscope is given by:$m = -\dfrac {v_{o}}{u_{o}} \left (1 + \dfrac {D}{f_{e}}\right )$Where $v_{o}$ is the image distance from the objective lens$D$ is the least distance of distinct vision$f_{e}$ is the focal length of eye lens.

Question 7

A simple microscope has a focal length of 5cm. The magnification at the least distance of distinct vision is

Accepted Answer

For single microscope,$m= 1+\dfrac{D}{f}$    $= 1+\dfrac{25}{5}$    $= 6$

Question 8

The power of plane mirror is ____

Accepted Answer

As radius of curvature of plane mirror is infinitySo, its focus $f = \dfrac {R}{2}$ is also infinity.Now, power $= \dfrac {1}{f} = \dfrac {1}{\infty} = 0$.

Question 9

In a compound microscope the focal length of objective lens.

Accepted Answer

As the  magnifying power of compound microscope is inversely proportional to the focal length of the objective lens. Hence , smaller is the focal length, higher is the magnifying power.Therefore, focal length of objective lens is less than the focal length of eyepiece lens.

Question 10

The focal length of a convex lens is 10 cm. The magnifying powers when it is used as a magnifying glass to form the image at (i) near point and (ii) far point are :

Accepted Answer

For the near point, magnification, $ {m}_{1} = 1 + \dfrac{D}{f} $For the far point, $ {m}_{2} = \dfrac{D}{f} $Thus, putting $D = 25$ and $f = 10$We get, $ {m}_{1} = 3.5$   and   ${m}_{2} =2.5 $

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