Ncert Solutions Physics Class 12th

9.15 For a concave mirror, the focal length, f < 0

When the object is placed on the left side of the mirror, the object distance, u is negative

For image distance v, we can write the lens formula as:

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$ (since u is negative) ……….(1)

The object lies between f and 2f, i.e. 2f < u < f

or, $\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$

or, $-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$

or, $\frac{1}{f}$ $-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}$ < 0 …………(2)

Using equation (1), we get

$\frac{1}{2f}<\frac{1}{v}$ < 0

Therefore, $\frac{1}{v}$ is negative, hence v is negative

$\frac{1}{2f}<\frac{1}{v}or$ 2f > v or –v > -2f . Hence, the image lies beyond 2f

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

Using equation (2), we can conclude that:

$\frac{1}{v}<0$ or v > 0

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

But we have u < 0, then

$\frac{1}{v}>\frac{1}{f}$ or v < f

$\mathrm{H}$ ence, the image formed is diminished and it is located between the focus (f) and the pole.

For a concave mirror, the focal length (f) is negative, f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative, u < 0

It is placed between the focus (f) and the pole, so f > u > 0

So, $\frac{1}{f}$ < $\frac{1}{u}$ < 0 or $\frac{1}{f}$ - $\frac{1}{u}$ < 0

For image distance v, we have the formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}$ < 0 or v > 0

The image is formed on the right side of the mirror, hence it is a virtual image.

For u < 0 and v > 0, we can write $\frac{1}{u}$ > $\frac{1}{v}$ or v > u

Magnification, m = $\frac{v}{u}$ > 1

Hence, the formed image is enlarged.

9.14 Focal length of the objective lens, $f_o$ = 15 m = 1500 cm

Focal length of the eyepiece, $f_e$ = 1.0 cm

The angular magnification of the telescope is given by,

m = $\frac{f_o}{f_e}$ = $\frac{1500}{1}$ = 1500

Hence the angular magnification of the telescope is 1500

Diameter of the moon, $d_m$ = 3.48 $*{10}^6$ m

Radius of the lunar orbit, $r_l$ = 3.8 $*{10}^8$ m

Let $d_1$ be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image. Hence,

$\frac{d_m}{r_l}$ = $\frac{d_1\mathrm{}}{f_o\mathrm{}}$

$d_1=\frac{d_m*f_o}{\mathrm{}{r}_l}$ = $\frac{3.48\mathrm{}*{10}^6*1500\mathrm{}}{3.8\mathrm{}*{10}^8}$ = 13.74 cm

Hence the diameter of the moon's image is 13.74 cm.

9.12 Focal length of the objective lens, $f_o$ = 8 mm = 0.8 cm

Focal length of the eyepiece, $f_e$ = 2.5 cm

Object distance for the objective lens, $u_o$ = - 9.0 mm = -0.9 cm

Least distance of distant vision, d = 25 cm

Image distance of the eyepiece, $v_e$ = -d = -25 cm

Object distance of the eyepiece = $u_e$

Using the lens formula, we can obtain the value of $u_e$ as:

$\frac{1}{v_e}$ - $\frac{1}{u_e}$ = $\frac{1}{f_e}$ or $\frac{1}{u_e}$ = $\frac{1}{v_e}-\frac{1}{f_e}$ = $\frac{1}{-25}-\frac{1}{2.5}$

$u_e=-2.27cm$

Using the lens formula, we can obtain the value of $v_o$ as:

$\frac{1}{v_o}$ - $\frac{1}{u_o}$ = $\frac{1}{f_o}$ or $\frac{1}{v_o}$ = $\frac{1}{u_o}+\frac{1}{f_o}$ = $\frac{1}{-0.9}+\frac{1}{0.8}$

$v_o=7.2cm$

The distance between the objective lens and the eyepiece = $\left|u_e\right|$ + $v_o$ = 2.27 + 7.2 = 9.47 cm

The magnifying power of microscope is calculated as:

m = $\frac{v_o}{\left|u_o\right|}$ ( 1 + $\frac{d}{f_e\mathrm{}}$ ) = $\frac{7.2}{0.9}$ ( 1 + $\frac{25}{2.5\mathrm{}}$ ) = 88

Hence the magnification of the microscope is 88.

9.11 Focal length of the objective lens, $f_1$ = 2.0 cm

Focal length of the eyepiece, $f_2$ = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, d' = 25 cm

Hence, image distance for the eyepiece, $v_2$ = - 25 cm

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{-25}-\frac{1}{6.25}$

$u_2=-5cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 5 = 10 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{10}-\frac{1}{2}$

$u_1=-2.5cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.5 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}$ ( 1 + $\frac{d\text{'}}{f_2\mathrm{}}$ ) = $\frac{10}{2.5}$ ( 1 + $\frac{25}{6.25}$ ) = 20

Hence the magnifying power of the microscope is 20

The final image is formed at infinity.

Hence, $v_2$ = $\infty$

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{\infty }-\frac{1}{6.25}$

$u_2=-6.25cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 6.25 = 8.75 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{8.75}-\frac{1}{2}$

$u_1=-2.59cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.59 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}*$ $\frac{d\text{'}}{\left|u_2\right|}$ = $\frac{8.75}{2.59}*\frac{25}{6.25}$ = 13.51

Hence the magnifying power of the microscope is 13.51.

9.9 Size of the object, $h_1$ = 3 cm

Object distance, u = - 14 cm

Focal length of the concave lens, f = - 21 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{-14}$ = $-\frac{1}{21}$

$\frac{1}{v}$ = $-\frac{1}{14}-\frac{1}{21}$ or v = $-\frac{42}{5}$ = - 8.4 cm

Hence, the image is formed on the other side of the lens, 8.4 cm away from the lens. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

m = $\frac{Imageheight\left(h_2\right)}{Objectheight\left(h_1\right)}$ = $\frac{v}{u}$

$\frac{h_2}{h_1}$ = $\frac{-8.4}{-14}$ or $h_2$ = 0.6 $*3$ = 1.8 cm

If the object is moved further away from the lens, then the virtual image will move towards the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.