Optics(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

OPTICS MCQs

Wave Optics, Ray Optics Refraction, Ray Optics Fundamentals, Ray Optics Curved Surface Refraction, Ray Optics Curved Mirrors

Total Questions : 111 | Page 10 of 12 pages

Question 91. A person sees his virtual image by holding a mirror very close to the face.
When he moves the mirror away from his face, the image becomes inverted.
What type of mirror he is using

Plane mirror
Convex mirror
Concave mirror
None of these

Discuss Question

Answer: Option C. -> Concave mirror
:
C
Plane mirror and convex mirror always forms erect images. Image formed by
concave mirror may be erect or inverted depending on position of object.

Question 92. Figure shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror

Is plane
Is convex
Is concave
May be any spherical mirror

Discuss Question

Answer: Option A. -> Is plane
:
A
Lets first of all think about the situation given to us. Two parallel incident rays give us two parallel reflected rays. It means that at both the points the normal must be parallel, only then this is possible. (Think from the point of view of the law of reflection followed by both the rays.)
Now, we know that, for a spherical mirror, the normal at any point passes through the center of curvature. So, if all the normals pass through a point, they are not parallel.
So, we can only have parallel normal in the case of a plane mirror. So our mirror here is a plane mirror.

Question 93. All of the following statements are correct except

The magnification produced by a convex mirror is always less than one
A virtual, erect, same-sized image can be obtained using a plane mirror
A virtual, erect, magnified image can be formed using a concave mirror
A real, inverted, same-sized image can be formed using a convex mirror

Discuss Question

Answer: Option D. -> A real, inverted, same-sized image can be formed using a convex mirror
:
D
Convex mirror always forms, virtual, erect and smaller image.

Question 94. 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

f2&I2
f&I4
3f4&I2
f&3I4

Discuss Question

Answer: Option B. -> f&I4
:
B
There will be no changein 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}

Question 95. 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

0
5 cm
2.5 cm
10 cm

Discuss Question

Answer: Option C. -> 2.5 cm
:
C
In the following ray diagram

Δ^{'} s, A B C

and CDE are symmetric

A Point Object O Is Placed On The Principal Axis Of A Convex...

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 96. 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

mx(m+1)2
mx(m−1)2
(m+1)2mx
(m−1)2mx

Discuss Question

Answer: Option A. -> mx(m+1)2
:
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 97. 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 3f2, 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 98. 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 99. 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]