Answer

Verified

384.6k+ views

**Hint:**First we have to draw the figure of both the mirrors and lens after that we use the simple rule of converging mirror and that is when we place object at center of the curvature of converging or concave mirror than image will produced at the center of the curvature with the same height by using this rule we will find where shall we place object.

**Formula used:**

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v = distance of image from the lens

u = distance of the object from the mirror

f = focal length of the mirror

**Complete answer:**

$\to $ Now assume that object is placed at a distance of y so that formula for the lens is given by

$\dfrac{1}{v}-\dfrac{1}{y}=\dfrac{1}{f}$

$\to $ Now substituting value of f for the lens we get

$\dfrac{1}{v}-\dfrac{1}{y}=\dfrac{1}{-20}....\left( 1 \right)$

$\to $In concave lens focal length will be negative (-ve).

$\to $Now suppose the image produced by the lens is at the center of the curvature in the concave mirror.

Now,

$\begin{align}

& c=2\times 10 \\

& c=20cm \\

\end{align}$

$\to $Now both lens and mirror are 5cm apart so the distance of the center of the curvature is

$\begin{align}

& =20-5 \\

& =15cm \\

\end{align}$

$\to $ Now if we assume that the image produced by the reflection is at the center of the curvature then the image produced by the reflection of the mirror is also at the center of the curvature.

$\to $Now it is again refract from point c so we can say that image distance from the lens is

$v=-15cm$

$\to $Now substitute value of the v in equation (1) we get

$\begin{align}

& \Rightarrow -\dfrac{1}{15}-\dfrac{1}{7}=-\dfrac{1}{20} \\

& \Rightarrow -\dfrac{1}{y}=-\dfrac{1}{20}+\dfrac{1}{15} \\

& \Rightarrow \dfrac{1}{y}=\dfrac{1}{20}-\dfrac{1}{15} \\

& \Rightarrow \dfrac{1}{y}=\dfrac{-5}{300} \\

& \Rightarrow \dfrac{1}{y}=-\dfrac{1}{60} \\

& \Rightarrow y=-60cm \\

\end{align}$

Therefore we can say that the object should be placed at 60cm from the left side of the lens to produce an image that coincides with the object.

**Hence the correct option is (A) 60cm away from the lens.**

**Note:**

When we are putting values from the left side of the lens all values should be taken with the negative sign. Otherwise it can lead us to wrong solution of this type of question

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Select the word that is correctly spelled a Twelveth class 10 english CBSE