Ray Optics Curved Surface Refraction(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

RAY OPTICS CURVED SURFACE REFRACTION MCQs

Total Questions : 30 | Page 1 of 3 pages

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

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

Discuss Question

Answer: Option A. ->

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

:
A

Question 2.

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
3

Discuss Question

Answer: Option C. -> 1.33
:
C

Question 3.

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]

Discuss Question

Answer: Option D. ->

:
D

Question 4.

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

Air and placed in air
Air and immersed in $L_{1}$
$L_{1}$ and immersed in $L_{2}$
$L_{2}$ and immersed in $L_{1}$

Discuss Question

Answer: Option D. ->

L_{2}

and immersed in

L_{1}

:
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 5.

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]

40 cm
20 cm
80 cm
30 cm

Discuss Question

Answer: Option A. -> 40 cm
:
A

Question 6.

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

[MP PMT 1994]

Discuss Question

Answer: Option C. ->

:
C

Question 7.

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

At one of the foci, virtual and double its size
At $\frac{3 f}{2}$ , real and inverted
At 2f, virtual and erect
None of these

Discuss Question

Answer: Option A. -> At one of the foci, virtual and double its size
:
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 8.

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

50 cm
30 cm
– 50 cm
– 30 cm

Discuss Question

Answer: Option D. -> – 30 cm
:
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 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

8.6 cm
6.2 cm
10 cm
4.4 cm

Discuss Question

Answer: Option D. -> 4.4 cm
:
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}

∴

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}

f = - \frac{40}{9}

or f = – 4.4 cm

A Concave Lens Forms The Image Of An Object Such That The Di...

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

10 cm
5 cm
20 cm
15 cm

Discuss Question

Answer: Option D. -> 15 cm
:
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 =