Geometrical Optics — Page 9 | NEET Physics

Q81

A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances x1 and x2 (x1 > x2) from the lens. The ratio of x1 and x2 is :-

A 3 : 1

B 2 : 1

C 5 : 3

D 4 : 3

Correct Answer

Option A

Solution

Magnification is 2 If image is real,

{x_1} = {{3f} \over 2}

If image is virtual,

{x_2} = {f \over 2}

{{{x_1}} \over {{x_2}}} = 3:1

Q82

Red light differs from blue light as they have :

A Different frequencies and same wavelengths

B Different frequencies and different wavelengths

C Same frequencies and different wavelengths

D Same frequencies and same wavelengths

Correct Answer

Option B

Solution

Red light and blue light have different wavelength and different frequency.

Q83

A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index

\dfrac{5}{3}

is poured inside the bucket, then the microscope has to be raised by

30 \mathrm{~cm}

to focus the object again. The height of the liquid in the bucket is :

A

50 \mathrm{~cm}

B

18 \mathrm{~cm}

C

75 \mathrm{~cm}

D

12 \mathrm{~cm}

Correct Answer

Option C

Solution

Shift $=\left(d-\dfrac{d}{\mu}\right)=30 \mathrm{~cm}$

\begin{aligned} & \Rightarrow d\left[1-\frac{1}{\frac{5}{3}}\right]=30 \\\\ & \Rightarrow d=\frac{30 \times 5}{2}=75 \mathrm{~cm} \end{aligned}

Q84

An object is placed at a distance of 12 cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm towards the stationary object. The distance by which the position of image would be shifted, will be

A 4 cm towards mirror

B 2 cm towards mirror

C 8 cm away from mirror

D 8 cm towards mirror

Correct Answer

Option D

Solution

$\therefore$ Shifting of image will be 8 cm towards mirror.

Q85

A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ________.

A

\dfrac{60}{7}

B 15

C 45

D 30

Correct Answer

Option B

Solution

Equivalent focal length

\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\ & =\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60} \\ & \mathrm{f}=-60 \mathrm{~cm} \end{aligned}

Lens formula

\begin{aligned} & \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{-20}=\frac{1}{-60} \\ & \mathrm{v}=-15 \mathrm{~cm} \end{aligned}

Q86

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :

A 22 mm

B 12 mm

C 33 mm

D 2 mm

Correct Answer

Option A

Solution

Case 1 : Near – point adjustment M.P =

{L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)

$\Rightarrow$ 375 =

{{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)

$\Rightarrow$ fe = 21.7 mm $\approx$ 22 mm Case-2 : If final image is at inifinity M.P =

{L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)

$\Rightarrow$ 375 =

{{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)

$\Rightarrow$ fe = 20 mm

Q87

For the thin convex lens, the radii of curvature are at

15 \mathrm{~cm}

and

30 \mathrm{~cm}

respectively. The focal length the lens is

20 \mathrm{~cm}

. The refractive index of the material is :

A 1.2

B 1.5

C 1.4

D 1.8

Correct Answer

Option B

Solution

To find the refractive index of the material of a thin convex lens, we can make use of the Lensmaker's Formula.

The Lensmaker's formula is given by:

\frac{1}{f} = \left( \frac{\mu - 1}{\mu} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

where $f$ is the focal length of the lens, $\mu$ is the refractive index of the material of the lens, $R_1$ is the radius of curvature of the first surface (convex side, positive), $R_2$ is the radius of curvature of the second surface (concave side, negative for convex lens).

Given in the question, the radii of curvature are $15 \, \mathrm{cm}$ and $30 \, \mathrm{cm}$ respectively, and the focal length $f = 20\, \mathrm{cm}$ .

It's important to pay attention to the signs of the radii of curvature according to the lens maker's convention.

For convex lenses, $R_1$ is positive and $R_2$ is negative; however, since the problem doesn't specify which curvature corresponds to which side in relation to the direction of light travel, we'll assume the light travels from left to right: thus, $R_1 = +15 \, \mathrm{cm}$ and $R_2 = -30 \, \mathrm{cm}$ .

Substituting the given values into the Lensmaker's Formula, we get:

\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} + \frac{1}{30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{2 + 1}{30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)

\frac{1}{20} = (\mu - 1) \left( \frac{1}{10} \right)

\frac{1}{20} = \frac{\mu - 1}{10}

Now, solve for $\mu$ :

\mu - 1 = \frac{10}{20}

\mu - 1 = 0.5

\mu = 1.5

Hence, the refractive index of the material of the lens is 1.5, which corresponds to Option B.

Q88

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is

2

cm. The focal lengths of the component lenses are :

A 10 cm, 12 cm

B 12 cm, 14 cm

C 16 cm, 18 cm

D 18 cm, 20cm

Correct Answer

Option D

Solution

For a convergent doublet of separated lens, we have

{1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}

...... (1) where d is separation between two lens, f1 and f2 are focal lengths of component lenses, f is resultant focal length.

Therefore, Eq. (1) becomes

{1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)

\Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20

\Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} = + 20

For f1 = 18 cm and f2 = 20 cm, the above equation satisfies.

Q89

If the refractive index of the material of a prism is

\cot \left(\dfrac{A}{2}\right)

, where

A

is the angle of prism then the angle of minimum deviation will be

A

\pi-2 \mathrm{~A}

B

\dfrac{\pi}{2}-2 \mathrm{~A}

C

\pi-\mathrm{A}

D

\dfrac{\pi}{2}-\mathrm{A}

Correct Answer

Option A

Solution

\begin{aligned} & \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\ & \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\ & \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\ & \delta_{\min }=\pi-2 \mathrm{~A} \end{aligned}%

Q90

A bi-convex lens has radius of curvature of both the surfaces same as

1 / 6 \mathrm{~cm}

. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides

\left(R_1 \neq R_2\right)

, without any change in lens power then possible combination of

R_1

and

R_2

is :

A

\dfrac{1}{3} \mathrm{~cm}

and

\dfrac{1}{7} \mathrm{~cm}

B

\dfrac{1}{5} \mathrm{~cm}

and

\dfrac{1}{7} \mathrm{~cm}

C

\dfrac{1}{3} \mathrm{~cm}

and

\dfrac{1}{3} \mathrm{~cm}

D

\dfrac{1}{6} \mathrm{~cm}

and

\dfrac{1}{9} \mathrm{~cm}

Correct Answer

Option B

Solution

To replace a bi-convex lens with another convex lens that has different radii of curvature on each side (i.e., $R_1 \neq R_2$ ), while maintaining the same lens power, the radii must satisfy a specific relationship.

For the current bi-convex lens, both radii of curvature are given as $\dfrac{1}{6} \, \text{cm}$ .

The lens formula for power is tied to the radii through: $\dfrac{1}{f_1} = (\mu - 1) \left( \dfrac{2}{R} \right)$ When this lens is replaced by a lens with different radii ( $R_1$ and $R_2$ ), the condition that has to be met is: $\dfrac{1}{f_2} = (\mu - 1) \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)$ For power equivalence, the expressions for $\dfrac{1}{f_1}$ and $\dfrac{1}{f_2}$ must be equal: $(\mu - 1) \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right) = (\mu - 1) \left( \dfrac{2}{R} \right)$ This simplifies to: $\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{2}{R}$ Given $R = \dfrac{1}{6} \, \text{cm}$ , it follows that: $\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{2}{\left(\dfrac{1}{6}\right)}$ Thus, simplifying the equation gives: $\dfrac{1}{R_1} + \dfrac{1}{R_2} = 12$ Therefore, any valid pair of radii $R_1$ and $R_2$ must satisfy this equation.