Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If in a plano-convex lens, the radius of curvature of the convex surface is 10 cm and the focal length of the lens is 30 cm, then the refractive index of the material of lens will be

[CPMT 1986; MNR 1988; MP PMT 2002; UPSEAT 2000]

1.5

1.66

1.33

D. 3

SOLUTION

Solution : C

Question 2

A convex lens of focal length f is placed somewhere in between an object and a screen. The distance between the object and the screen is x. If the numerical value of the magnification produced by the lens is m, then the focal length of the lens is

\frac{m x}{(m + 1)^{2}}

\frac{m x}{(m - 1)^{2}}

\frac{(m + 1)^{2}}{m} x

\frac{(m - 1)^{2}}{m} x

SOLUTION

Solution : A

Question 3

The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is 1.5. The focal length will be

[MP PMT 1989]

A. 40 cm

20 cm

80 cm

30 cm

SOLUTION

Solution : A

Question 4

The graph shows how the magnification m produced by a convex thin lens varies with image distance v. What was the focal length of the used
[DPMT 1995]

SOLUTION

Solution : D

Question 5

The graph between the lateral magnification (m) produced by a lens and the distance of the image (v) is given by

[MP PMT 1994]

SOLUTION

Solution : C

Question 6

A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids $L_{1}$ and $L_{2}$ having refractive indices $n_{1}$ and $n_{2}$ respectively $(n_{2} > n_{1} > 1)$ . The lens will diverge a parallel beam of light if it is filled with

A. Air and placed in air

B. Air and immersed in

L_{1}

L_{1}

and immersed in

L_{2}

L_{2}

and immersed in

L_{1}

SOLUTION

Solution : D

$\frac{1}{f} = (\frac{n_{2}}{n_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ Where $n_{2}$ and $n_{1}$ are the refractive indices of the material of the lens
and of the surroundings respectively. For a double concave lens,

$(\frac{1}{R_{1}} - \frac{1}{R_{2}})$ is always negative.

Hence f is negative only when $n_{2} > n_{1}$

Question 7

In order to obtain a real image of magnification 2 using a converging lens of focal length 20 cm, where should an object be placed

A. 50 cm

B. 30 cm

C. – 50 cm

D. – 30 cm

SOLUTION

Solution : D

For real image m=-2

$∴$ M = $\frac{f}{u + f} \Rightarrow - 2 = \frac{f}{u + f} = \frac{20}{u + 20} \Rightarrow$ u=-30 cm.

Question 8

An object is placed at a distance of $\frac{f}{2}$ from a convex lens. The image will be

A. At one of the foci, virtual and double its size

B. At

\frac{3 f}{2}

, real and inverted

C. At 2f, virtual and erect

D. None of these

SOLUTION

Solution : A

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (Given $u = \frac{- f}{2}$ )
$\Rightarrow \frac{1}{f} = \frac{1}{v} + (\frac{1}{\frac{f}{2}}) \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{2}{f}$
$\Rightarrow \frac{1}{v} = \frac{- 1}{f}$ and $m = \frac{v}{u} = \frac{f}{\frac{f}{2}}$ = 2
So virtual at the focus and of double size.

Question 9

A concave lens forms the image of an object such that the distance between the object and image is 10 cm and the magnification proceed is $\frac{1}{4}$ . The focal length of the lens will be

A. 8.6 cm

B. 6.2 cm

C. 10 cm

D. 4.4 cm

SOLUTION

Solution : D

Concave lens forms the virtual image of a real object. So let
u = -4x and v = -x then 3x = 10 cm
or $x = \frac{10}{3}$ cm
$∴$ $u = - \frac{40}{3} c m$
And $v = - \frac{10}{3} c m$
Substituting in $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
We get $\frac{1}{f} = \frac{- 3}{10} + \frac{3}{40}$
or $f = - \frac{40}{9}$
or f = – 4.4 cm

Question 10

A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object the image is formed at infinity. The thickness t is

A. 10 cm

B. 5 cm

C. 20 cm

D. 15 cm

SOLUTION

Solution : D

Image will be formed at infinity if object is placed at focus of the lens i.e., at 20 cm from thelens. Hence,
Shift = $25 - 20 = (1 - \frac{1}{μ}) t$
or $5 = (1 - \frac{1}{1.5}) t$
or $t = \frac{5 \times 1.5}{0.5} = 15 c m$

Question 11

\frac{m x}{(m + 1)^{2}}

\frac{m x}{(m - 1)^{2}}

\frac{(m + 1)^{2}}{m} x

\frac{(m - 1)^{2}}{m} x

SOLUTION

Solution : A

$\frac{v}{- u} = - m; v + u = x \Rightarrow u = \frac{x}{1 + m}$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow f = \frac{m x}{(m + 1)^{2}}$

Question 12

An object is placed at a distance of $\frac{f}{2}$ from a convex lens. The image will be formed:

At $- f$ , virtual and double its size.

At $f$ , real and inverted.

At $- 2 f$ , virtual and erect.

D. At

2 f

, real and erect.

SOLUTION

Solution : A

Given $u = \frac{- f}{2}$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{v} + (\frac{1}{\frac{f}{2}}) \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{2}{f}$
$\Rightarrow \frac{1}{v} = \frac{- 1}{f}$
So $v = - f$ i.e., the image will be formed at focus of the lens on the same side as that of the object.
Now magnification $m = \frac{v}{u} = \frac{- f}{\frac{- f}{2}} = + 2$
So the image formed will be virtual with magnification +2.

Question 13

A point object O is placed on the principal axis of a convex lens of focal length 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. If the eye is placed 60 cm to the right of the lens at a distance h below the principal axis, then the maximum value of h to see the image will be

A. 0

B. 5 cm

C. 2.5 cm

D. 10 cm

SOLUTION

Solution : C

In the following ray diagram $Δ^{'} s, A B C$ and CDE are symmetric

So, $\frac{A B}{B C} = \frac{D E}{C D} \Rightarrow \frac{5}{40} = \frac{h}{20} \Rightarrow h = 2.5 c m$

Question 14

The object distance u, the image distance v and the magnification m in a lens follow certain linear relations. These are

\frac{1}{u}

versus

\frac{1}{v}

B. m versus u

C. u versus v

D. m versus

\frac{1}{u}

SOLUTION

Solution : A

For a lens $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} . . . . (i)$
Also $m = \frac{f - v}{f} = 1 - \frac{v}{f} \Rightarrow m = (- \frac{1}{f}) v + 1 . . . (i i)$
On comparing equations (i) and (ii) with y =mx +c.
It is clear that relationship between $\frac{1}{v} v s \frac{1}{u}$ and m vs v is linear.

Question 15

A thin lens has focal length f and aperture radius r. It forms an image of intensity I. Now the central part of the aperture upto $\frac{r}{2}$ is blacked by an opaque material. The focal length and intensity of image will be

\frac{f}{2} & \frac{I}{2}

f & \frac{I}{4}

\frac{3 f}{4} & \frac{I}{2}

f & \frac{3 I}{4}

SOLUTION

Solution : B

There will be no change in focal length. $f^{1} = f$ But Intensity $I \propto A$
But Intensity $I \propto A$
$\frac{I^{1}}{I} = \frac{r_{1}^{2}}{r^{2}} = \frac{1}{4} ∵ r_{1} = \frac{r}{2} I^{1} = \frac{I}{4}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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