Practice for u, a help for me 2

Some problems from geometric optics

Q1) Two mirrors are inclined at an angle $\alpha$ .Light ray is inclined parallel to one og the mirrors.The ray will start retracing its path after third reflection if $\alpha=$ .The figure is above

Q2) A point object is kept in front of a plane mirror. The plane mirror is doing SHM of amplitude $2~ cm$ .The plane mirror moves along the $x-axis$ and $x-axis$ is normal to the mirror.The amplitude of the mirror is such that the object is always in front of mirror.The amplitude of SHM of the image is

Q3) A person's eye is at height of $1.5~m$ . He stands in front of a $0.3~m$ long plane mirror which is $0.8~m$ above the ground. the length of the image he sees of himself is

Q4) A person is standing in a room of width $200~cm$ . A plane mirror of vertical length $10~cm$ is fixed on a wall in front og the person. The person looks into the mirror from distance $50~cm$ .How much width of the wall behind him will he able to see (assume that he uses the full mirror).

Q5) An unnumbered wall clock shows time $04:25:37$ , where $1st$ terms represent hours, $2nd$ represent minutes and $3rd$ represents seconds. What time will its image in a plane mirror show..

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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image2

image3

For problem 1 and 2 check the first image, I'll explain problem 3(answer for 1 is 30 degrees)

At first and a ray from A to top end of mirror after reflection it will hit C(C will be at distance .8m from A), Now similarly send a ray to bottom end, after reflection it will hit E(E is at a distance 1.4m from A) and he can see himself from C to E which equals 0.6m.

Similarly in problem 4 send ray to ends of mirror and apply simple geometr(our answer = 2x + 10 = 50)

For problem 5 answer is equivalent when you put mirror on axis containing numbers 6 and 12.

Pro tip: In mostly all plane mirror problems you just need to send rays at extremities of mirror and apply simple geometry.

Krishna Sharma - 6 years, 2 months ago

In the 1st ques. Ans is 30degree &in 4th 50cm

Munendra Singh - 3 years, 10 months ago

1) 30 2) 4 cm 3) 0.6 m 4) 50 cm 5) 7:34:23

Rohit Shah - 6 years, 2 months ago

30 degrees.

1) 30 degrees and 5) 7:34:23.

Tanishq Varshney - 6 years, 2 months ago

You are correct I did a silly mistake in both. I realized later

@Rohit Shah – can u explain me the 5) one

@Tanishq Varshney – Just draw the clock and reflect . Or you could subtract 12 hours from the given time(its effectively the same thing)

@Rohit Shah – then how is 25 replaced by 34 , why isnt it 35

@Tanishq Varshney – Another simple way which prevents confusion is that you subtract the given time from $11: 59 : 60$ component wise. So here $11:59:60 - 04:25:37 = 07:34:23$ which is the required answer.

Sudeep Salgia - 6 years, 2 months ago

@Sudeep Salgia – Yeah, this is indeed the best way out .

A Former Brilliant Member - 6 years, 2 months ago

@Tanishq Varshney – Because the seconds add up to give an extra minute. 37 + 23 is the extra minute. You could have used a similar reasoning that 4 + 7 is not 12 but we actually have got an extra hour by adding 25 and 35 mins. So sudeep has suggested a fast way of subtracting 12 hours

@Deepanshu Gupta @Rohit Shah @Azhaghu Roopesh M @Raghav Vaidyanathan ,plz post a solution for the above problems

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$