Solution of HC Verma : Concept of Physics – I : Chapter 18 Geometrical Optics

5/21/2012

100 Comments

Solution of HC Verma : Concept of Physics – I : Chapter 18 Geometrical Optics

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100 Comments

NISHARG GOLASH link

5/5/2013 10:46:25 pm

Thanks a lot for providing the solutions, being a teacher I was searching for them for a long time.

http://www.goodiesforbuddies.com/

theyatin

5/6/2013 12:33:25 am

dear,
a teacher is first a student then a teacher. . .
so first learn and then teach each time you teach you learn more. .. :)

jailal

5/18/2013 02:52:38 pm

explain ques no.14

theyatin

5/19/2013 12:29:15 am

dear,
dear sin i/sin r= velocity of light / speed of light inside medium
distance travelled by light inside slab is cos 30=base/hypotnese
that is x/b=cos 30
velocity =distance/time
thus we can calculate time taken as we have distance as well as velocity.

PRATIBHA PATEL

5/18/2013 07:47:46 pm

why refracted angle should be 45 in ques 24

theyatin

5/19/2013 12:32:28 am

dear,
because angle r is 45' thus opposite angle of bisecting lines is 45'

PRATIBHA PATEL

5/20/2013 10:55:42 pm

why angle 45 is needed for the point to be visible

theyatin

5/21/2013 10:49:02 pm

dear,
as it is given in drawing of question in book its 45'

anurag

6/28/2013 04:18:35 pm

because corner is seen by eye and height and diameter are both equal,i.e., 30 cm.
so tan i=30/30
by this tan 45=1;
hence i=45;

ram

6/12/2013 03:07:08 am

some calculations are wrong in Q41 please check it

theyatin

6/15/2013 06:14:32 pm

dear calculations are correct.
as 1/11=9.09 that can be written as 9.1

divyansh meena

7/15/2013 03:10:03 pm

sir i am unable to understand that why can't we use equivalent focal lenth in q no. 41 and get the answer in one step instead of first solving for part A and then for part B

theyatin

7/16/2013 02:59:03 am

dear,
in this numerical we need to find focal length of two half convex lens which are combined to make a thick lens.
thus combined focal length would be the answer.
we need to solve the numerical as per conditions given in the question.

Rameeza

7/23/2013 03:32:00 am

in ques no. 61 .. since image is formed at same place then for the mirror the object must be at centre of curvature and it would act as image distance for the lens ( -15 cm ) ..... instead of solving in such a long process.. can't we do this..?

teyatin

7/25/2013 12:13:20 am

dear,
there is always a short formula but short formulas are not standard and thus can not be used everywhere so let it be.

Rameeza

7/25/2013 02:24:58 am

Sir, will u give me the explanation of ques 73 nd 74 ?

theyatin

7/30/2013 01:19:34 am

dear 74
first we need to find distance then we need to find velocity that is time derivative of distance.
as object is moving towards mirror so the image would move away from mirror so distance between velocity would be considered by
difference of velocities of both distances. as the both distances are inverse to each other.

murli

8/14/2013 07:53:40 pm

teacher my question is that whether we need to study hc verma for BITSAT and JEE enterance

theyatin

8/16/2013 12:44:31 am

dear,
hc verma book improves your conceptual ability. . .
just practice and improve you basics.

murli

8/17/2013 03:31:27 am

could you please explain me Q NO. 76,77,78

mansi link

8/17/2013 09:42:36 pm

plz explain Q2 short questions

theyatin

8/18/2013 05:25:32 am

dear,
light deviates only when it enters denser medium from rare medium.
as in case of prism.
when prism is dipped in a solution of same refractive index as of prism light will not deviate.

another case is when light enters at such an angle that it is reflected from the third dace internally in prism and leave the prism from the opposite face at the same angle it entered the prism.
it is rare case.

Rameeza

8/18/2013 06:19:37 pm

Sir..73rd plz... !! :-)

Anindita

8/20/2013 04:16:57 pm

Sir, does d pdf contain the exercises solved, i mean the part after objective 2..?
I am not able to see it.. can u plz solve qstn no 24 from the exercise..

Anindita

8/20/2013 04:19:04 pm

I can only see the solution of the short qstns..where can i get the solution of the exercise.?

theyatin

8/23/2013 03:11:52 am

dear,
we regret for technical problem now all answers are available please look at the pdf now. . .

Anindita

8/24/2013 04:58:42 pm

Thankyou sir..!! :)

narayanhari

9/21/2013 03:35:43 pm

Qno:2 exercises

theyatin

9/23/2013 01:12:37 am

dear,
here v is distance of image thus taken as negative
as ratio of distance of image from mirror is equal to ratio of hieghts of images.
-v/u=h2/h1
we have v, h1 and h2
hence we can find u
thus from
1/v+1/u=1/f
we can find focal length

anhu nirankari

9/24/2013 09:11:22 pm

thanks sir my physics concpet really improve now i enjoy this

anshu nirankari

9/24/2013 09:13:12 pm

thanks you sir i enjoy physics

Anwesh Nayak

9/27/2013 01:27:02 am

Sir Chapter Geometrical Optics EXAMPLE Q 27..
Sir please ..

Anwesh Nayak

9/27/2013 10:57:50 am

Sir i coudn't get that refraction thing!! it says image was formed but by without glass slab or with glass slab???
and what for the image distance is taken as 9.5!! sir plzz do look after this example given in the book!

theyatin

9/29/2013 04:03:00 am

dear,
refraction means bending of light when medium of light is changed thus
image is made with glas slab.
and then without glass slab to observe change in path.

David

10/3/2013 01:32:39 am

How did you calculate time taken in Q.no.14

theyatin

10/4/2013 01:24:42 am

dear,
time taken =distance / velocity
velocity=c/ undrt 2 c=speed of light
distance = 2/undrt3

shabd

10/11/2013 03:56:12 pm

sir, plz give the solution of Q.17 of objective 1

Shivam Pd.

10/15/2013 06:10:01 pm

In Quest 22,formula is d(real)(1-U)
How come we used + instead of -ve,thanks

shabd

10/19/2013 03:24:10 am

sir, plz give the solution of
Q.17 of objective 1

shabd

10/20/2013 07:35:01 pm

plz explain Q.13(a) of exercise

shabd

10/20/2013 07:46:48 pm

plz give the solution of Q.15 of exercise

shabd

10/22/2013 03:34:48 pm

plz explain Q.16 of exercises

shabd

10/22/2013 03:36:41 pm

plz give the solution of Q.19 of exercises

shreya

10/22/2013 08:10:36 pm

sir, a ray of light falls on a surface of a spherical glass paper weight making an angle alpha with the normal and is refracted in the medium at an angle beta .the angle of deviation of the emergent ray from the dirction of the incident ray is.

shreya

10/22/2013 08:14:45 pm

sir, a glass concave lens is placed in which it behaves like a convergent lens .if the refractive indices of glass and liquid with respect to air are amug and amul respectively then

shreya

10/22/2013 08:19:17 pm

sir tthe diameter of aperture of a plano convex lens is 6cm and its maximum thickness is3mm .if the velocity of light in the material of lens is 2*10m/s its focal length is

shabd

10/23/2013 02:54:35 pm

plz give the solution my previous Q.s

bhavya

11/3/2013 01:43:10 am

in ques 8 y r v nt taking the case of magnification with negative sign

shabd

11/10/2013 05:36:19 pm

plz explain me Q.42 of exercises

theyatin

11/18/2013 01:36:11 am

i will tell you later.

dave

11/18/2013 01:41:56 am

i wrote that above comment since so many questions were being asked and bhavya we are taking the -ve sign into consideration but it is getting cancelled in the next step :
- v/u = -h2/h1

dave

11/18/2013 01:54:29 am

oh typing did not see that , bhavya that will be h2/h1=-v/u and u = - 25 therefore - v/u = -v/-25 = v/25 thus it is being mutually cancelled

Kshitiz

12/24/2013 09:17:07 pm

I AM NOT GETTING QUESTION NO.33
PLEASE HELP ME.
I WOULD BE GLAD IF SOMEONE HELP ME WITH THIS QUESTION

arpit

11/9/2015 05:58:57 am

Hope u got the second part on ur own by the formula sin i = 1/u then by converting into tan .
Then for the previous part first find the apparent depth by the formula d'/d=u2/u1 then with the height and the given radius get the angle and extend the light rays .Then by tan you can get the radius.

taniya sharma link

1/24/2014 03:03:56 pm

in ques 5 please explain the sign convention?

tiko

2/4/2014 11:21:52 pm

how can i download the problem only not the full books

theyatin

2/5/2014 09:27:12 pm

select
copy
paste in word file

tiko

2/4/2014 11:23:14 pm

how can i download the problem only not the full books

kashyap

2/14/2014 03:06:06 pm

My goodness..where was this website all this time..how could I overlook this..tremendous effort, solving the problems and then replying to the comments. Sir, I salute the awesomeness!! This website goes straight into my bookmarks.

ash link

4/15/2014 12:37:12 am

plz explain Q.no-32 nd 22

ggggggggukkkkkkkk

7/5/2014 04:57:20 pm

4/15/2014 06:11:53 pm

Q22 please explain properly .. the real depth is 3h/2 and not just h..then how come u used h and solved ?

Anmol

5/8/2014 03:55:09 pm

Sir, please explain q.no. 65 when both images are to be formed at same place?????

anjali

5/19/2014 11:57:19 pm

Sir how to solve question 11

praveen

11/7/2014 10:18:54 pm

Panikar

5/28/2014 04:46:24 pm

sir can i knw the procedure to solve ques no.12?

mirjaik

5/31/2014 12:28:10 am

md javed

6/16/2014 09:52:33 am

Sir how to solve 7

ABHINAV

6/17/2014 05:21:11 am

Plz explain ques 75 2nd part

ANAMITRA

8/17/2014 02:39:57 am

Can you please explain the 2nd reflection part of question no.13.

abhay

10/9/2014 07:16:05 pm

explain question no 22 of gemotrical optics

soup

10/10/2014 02:30:27 pm

sir, in question 79,,,i am not able to get fbd for the block m.
? thank u

levine

12/16/2014 12:02:56 am

Can you please explain the 15th question? Why is it (1+6.235)?

manishh

1/19/2015 06:18:10 pm

Sir, please explain question - 69

megha mishra

3/31/2015 10:42:44 pm

solve the questions from 6 to 15

kiara

4/10/2015 03:37:29 am

can u plz explain the 79 question????

Akash

5/26/2015 04:50:59 pm

47 part 2

Ram

9/8/2015 10:08:29 am

What is the correct nd genuine reason fo taking r=45° in 24 question.plz ans fast is any one can

Divanshu

9/25/2015 10:29:03 am

please explain Q15

tanya

10/1/2015 11:22:59 am

sir, in ques 61 ,we say that the placement of the object will be such that it will be on the center of curvature of the mirror then its distance from the lens will be taken as image distance, which means we are considering it as an image ,now we find the object distance from the LENS ..therefore at the end the object is NOT at the center of curvature of the mirror ??? why is this so ???

shiva

10/16/2015 02:11:05 pm

Hey yrr acha hai

arpit

11/9/2015 05:48:53 am

pls explain q43

arpit

11/10/2015 06:43:40 pm

in Q47 why cant we just obtain the effective focus and put the needle there so that image forms on itself

rAdha

12/23/2015 10:07:07 pm

its good to have

radha

12/23/2015 10:28:30 pm

i want 27 to58 all questions solutions

radha

12/24/2015 05:09:26 pm

want geometric optics solutions

Sanjana link

12/26/2015 06:02:14 am

Can you please explain me the 9th sum..
Thank you

Armaan

2/2/2016 01:25:43 am

Sir, can you explain Q) 7 from objective 2

science

2/8/2016 09:11:08 am

Where is the solution of question 47 b part?

Sayan

3/3/2016 02:35:45 am

Why in question no 24 d ray has struck d liquid beyond d mid point... Why not bear it??????

Prannoy Meht

3/3/2016 09:24:14 pm

There is an error in Q41. ug-ua = 0.5 not 1.5

joseph charles

9/9/2016 08:42:57 am

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ritwik

11/5/2016 07:56:34 am

in que. 40 there is written (5-15) where does 15 comes from

Abhinav link

11/25/2016 07:18:46 am

Find the position of object placed in front of a concave mirror of 15cm focal length so as to get an image magnified 3 times what is the answer with solution you should give in 3 minutes the answer with solution

Shobhit

12/1/2016 11:03:13 am

Respected Sir, I need an explanation of ques 47, part (b). I'll be very glad to welcome the reply as soon as possible.

ved prakash

12/11/2016 08:58:21 am

Ananya

11/26/2017 10:04:55 am

Can you please explain Q.no. 10?

Solution of HC Verma : Concept of Physics – I : Chapter 18 Geometrical Optics

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