## Applying Lenz’s Law

Nature abhors a change of flux.

D. J. Griffiths’ (2013) genius re-statement of Lenz’s Law, modelled on Aristotle’s historically influential but now debunked aphorism that ‘Nature abhors a vacuum’

A student recently asked for help with this AQA A-level Physics multiple choice question:

This question is, of course, about Lenz’s Law of Electromagnetic Induction. The law can be stated easily enough: ‘An induced current will flow in a direction so that it opposes the change producing it.’ However, it can be hard for students to learn how to apply it.

What follows is my suggested explanatory sequence.

### Step 1: simplify the diagram using the ‘dot and cross’ convention

When the switch is closed, a current I begins to flow in coil P. We can assume that I starts at zero and increases to a maximum value in a very small but not negligible period of time.

### Step 2: consider the magnetic field produced by P

You can read more about a simple method of deducing the direction of the magnetic field produced by a coil or a solenoid here.

### Step 3: apply Faraday’s Law to coil Q

Since Q is experiencing a change in magnetic flux, then an induced current will flow through it.

### Step 4: apply Lenz’s Law to coil Q

The current in coil Q must flow in such a direction so that it opposes the change producing it.

Since P is producing an increasing magnetic flux through Q, then the current in Q must flow in such a way so that it tries to prevent the increase in magnetic flux which is inducing it. The direction of the magnetic field BQ produced by Q must therefore be opposite to the direction of the magnetic field produced by P.

### Step 5: consider the polarity of the magnetic fields of P and Q

We can see the magnetic field lines of coil P produce a north magnetic field on its right hand side. The magnetic field of Q will produce a north magnetic field on its left hand side. Coil P will therefore push coil Q to the right.

It follows that we can eliminate options A and C from the question.

### Step 6: What happens when the magnetic field of P reaches its steady value?

Because the magnetic field produced by coil P has how reached its steady maximum value, this means that the magnetic flux through coil Q also has a constant, unchanging value. Since there is no change in magnetic flux, then this means that no emf is induced across the coil so no induced current flows. Since Q does not have a magnetic field it follows that there is no magnetic interaction between them.

The answer to the question must therefore be D.

### Step 7: check student understanding

For the alternative question, the correct answer of C can be explained by going through a process similar to the one outlined above.

• When the switch is opened, the magnetic flux through Y begins to decrease.
• A changing magnetic flux through Y induces current flow.
• Lenz’s Law predicts that the direction of this current is such that it opposes the change producing it.
• The current through Y will therefore be in the same direction as the current through X to produce a magnetic field in the same direction.
• The coils will attract each other.
• Eventually, the magnetic flux produced by coil X drops to a constant value of zero.
• Since there is no change in magnetic flux through Y, there is no induced current flow through Y and hence no magnetic field.
• There is no magnetic interaction between X and Y and therefore the force on Y is zero.

### Conclusion

I hope teachers find this detailed analysis of a Lenz’s Law question useful! As in much of A-level Physics, the devil is not in the detail but rather in the application of the detail. Students who encounter more examples will have a more secure understanding.

### Reference

Griffiths, David (2013). Introduction to Electrodynamics. p. 315.

## Dual coding change of momentum

Rosencrantz (an anguished cry): CONSISTENCY IS ALL I ASK!

Tom Stoppard, Rosencrantz and Guildenstern Are Dead (1966)

I think that dual coding techniques can be extremely helpful in helping students understand the concept of change of momentum.

To engage our students’ physical intuitions, let’s consider a question like: Which would hurt more — being hit by a sandbag or being hit by a rubber ball?

Let’s assume that the sandbag and rubber ball have the same mass m and are travelling at the same initial velocity u. We choose ‘u‘ because it’s the initial velocity and we take ‘v‘ as the final velocity: a very subtle piece of dual coding that can reap rewards if applied consistently — pace Rosencrantz(!) — over a range of disparate examples.

To analyse this problem, let’s use the momentum version of Newton’s Second Law of Motion.

We will use the change = final – initial convention (‘Consistency is all I ask!’)). The initial momentum is pi and the final momentum is pf.

Now let’s work out the change in momentum in each case. We will assume that each item is dropped so that it impacts vertically on a horizontal surface. The velocity just before it hits is u so its initial momentum pi is given by pi = mu; its final velocity is v so its final momentum pf is given by pf = mv. The sandbag does not rebound, so its final velocity v is zero.

The rubber ball rebounds from the surface with a velocity v (we have shown that v < u so we are not assuming a perfectly elastic collision).

We will use the down-is-positive convention so that u is positive and the downward momentum pi are positive in both cases. However, the velocity v of the ball is negative so the momentum pf = mv is negative (upwards).

To add vectors, we simply put them ‘nose to tail’. However, in this case, we need to subtract the vectors, not add them. To do this, we use the operation pf + (-pi,). In other words, we put the vector pf nose to tail with minus pi, or with a vector pointing in the opposite direction to the original vector pi. These are shown in the table.

We can see that the change in momentum Δp is larger in the case of the rubber ball.

Applying Newton Second Law that force = change in momentum / change in time then (assuming the time of each interaction is the same) then we can conclude that the (upward) force exerted by the surface on the ball is larger than the force exerted by the surface on the sandbag.

From Newton’s Third Law (that if an object A exerts a force on object B, then object B exerts an equal and opposite force on object A), we can also conclude that the rubber exerts a larger downward force on the surface. This implies that, if the ball hit (say) your hand, then it would hurt more than the sandbag.

Considering change of momentum problems like this helps students answer questions such as the one shown below:

We can discard options C and D since the change of momentum shown is in the wrong direction: the vertical component of momentum will remain unchanged.

A and B show changes of momentum of the same magnitude in the horizontal direction. However, if we take the horizontal component of the initial momentum as positive then the change of momentum on the gas particle must be negative; this implies that the correct answer is B.

Note also that diagram B shows the pf + (-pi) operation outlined above, with the arrow showing minus pi shown in red (added to the original exam question).

## Gravitational potential: the ‘bottom of a hole’ perspective

April 20, 2112: The sky is flat, the land is flat, and they meet in a circle at infinity. No star shows but the big one, a little bigger than it shows through most of the [asteroid] Belt, but dimmed to red, like the sky. It’s the bottom of a hole, and I must have been crazy to risk it. […] The stars are gone, and the land around me makes no sense. Now I know why they call planet dwellers ‘flatlanders’. I feel like a gnat on a table. I’m sitting here shaking, afraid to step outside. […] I’M AT THE BOTTOM OF A LOUSY HOLE!

Larry Niven, ‘At The Bottom of a Hole’ (1966)

Redish and Kuo (2015: 586) suggest that tapping into our students’ innate physical intuitions can be a very productive teaching strategy. For example, Redish observed some physics instructors teaching non-physics majors how to interpret a potential energy U against the separation r between particles graph (diagram 8(a) below).

The students were finding it difficult to answer the question of whether the particles would attract or repel each other when they had energy E and were at a separation of C. Redish noted that the instructors advised the students to consider the derivative of the curve at C (diagram 8(b) above) and, since it had a positive gradient, to surmise that the force between the particles would therefore be attractive since F=-dU/dr. Redish suggested:

A more effective approach for this population might be to begin with an embodied analogy and implicitly supporting epistemologies valuing physical intuition. Start with treating a potential energy curve as a track or hill and, using the analogy of gravitational potential energy, then place a ball on the hill as shown in Fig. 8c.

Redish and Kuo (2015)

Which way would the ball roll in 8(c) roll? Redish said that the students had no problem deducing that the particles would exert an attractive force on each other at C (and a repulsive force when their energy is E at the smaller value of r) after using this analogy.

### Using students’ physical intuitions to help understand gravitational potential

The episode outlined above reminded me of a science fiction story by Larry Niven that I had read many years ago. In ‘At the Bottom of a Hole’, Niven imagined what landing on a planet would feel like to a ‘Belter’; that is to say, to a human being who had spent their entire life navigating between the small worlds of the asteroid Belt: small planetoid-sized worlds whose shallow gravitational fields required only a low-intensity burn for a spaceship to slip free of their influence forever. An extract from the story is quoted as an introduction to this post: in essence, the ‘Belter’ who has lived his life voyaging between the low mass and low gravity worldlets of the asteroid belt finds it emotionally and psychologically disturbing to find himself at the bottom of a deep gravitational hole.

### Gravitational Fields are always ‘holes’

Gravitational fields are always holes (unlike electric fields, of course, which can be either ‘holes’ or ‘mountains’; this may well form the basis of a later post).

The mass of the Earth produces a much deeper gravitational hole than the much smaller mass of an asteroid.

As a consequence, a spaceship near the Earth’s surface (A) needs to burn a lot more fuel (i.e. do a lot more work) to completely escape the gravitational influence of the Earth (B) then a spaceship near to the surface of an asteroid. The spaceship closer to the asteroid (C) needs a much smaller burn to completely escape its gravitational influence (D).

To a mature space-faring civilisation, living on the surface of a planet could well be likened (and seem as eccentric) as living at the bottom of a spectacularly deep hole.

### Gravitational potential

The gravitational potential of an object of mass M is given by:

Note that the magnitude of V gets larger as r decreases. This allows us to represent a gravitational field in terms of equipotential lines (dotted on the diagram below) as well as field lines (solid).

### Modelling gravitational potential as a three dimensional hole

We can engage our own and our students’ physical intuitions by picturing the equipotential lines as being contour lines indicating the depth of a three dimensional hole.

An object represented as a ball at position A will not tend to roll down into the hole since there is no discernible downhill ‘slope’ at A; in effect, as r tends towards infinity then the object is beyond the effects of M’s gravity. A position outside the gravitational field of a massive object has a gravitational potential of zero.

Let’s think about what happens as r decreases until the object is at B. Here we can intuitively surmise that it will experience a small force tending to make it fall deeper into the hole. How much work will the gravitational field have done moving an object from infinity to this position? The answer is, of course, 0.5 MJ for each kilogram of mass.

How much work will be done by the gravitational field moving the object from B to C? The answer is again an extra 0.5 MJ/kg but note that this happens over a much smaller change in r than before because the gravitational field is becoming more intense. Again, we can intuit that the object will experience a stronger gravitational force at C than at B.

We can go on to argue that a similar pattern of behaviour will also occur at D and E.

But the real value of this representation is, in my opinion, helping students understand how much energy a body needs to escape the influence of a gravitational field.

If we start at B, we would have to do 0.5 MJ/kg of work on it to make it escape. In other words, it needs 0.5 MJ/kg to climb out of the hole.

If we started at C, then we would need 1.0 MJ/kg; and D, 1.5 MJ/kg and so on.

If we were considering a spacecraft operating in the vacuum of space, then transferring 2.0 MJ/kg of kinetic energy would allow ot to completely escape the gravitational influence of M; or, in other words, to reach a value of r such that its gravitational potential is zero.

Near the Earth’s surface where r = 6.38 x 106 m, the gravitational potential can be calculated as follows:

That is to say, a body would need to gain 64.4 MJ of kinetic energy for each kilogram of its mass to completely escape from the influence of the Earth’s gravity.

We can therefore calculate the escape velocity for a body near the Earth surface as follows:

As I mentioned above, I think the real power of this way of tapping into physical intuition for understanding fields comes when we use it to represent electric fields. I will cover that in a later post.

#### Reference

Redish, E. F., & Kuo, E. (2015). Language of physics, language of math: Disciplinary culture and dynamic epistemologyScience & Education24, 561-590.

## Just a moment . . .

Many years ago, I was taught this compact and intuitive convention to show turning moments. I think it should be more widely known, as it not only is concise and powerful, but also meets the criterion of being an effective form of dual coding which is helpful for both GCSE and A-level Physics students.

Let’s look at an example question.

Let’s start by ‘annotating the hell’ out of the diagram.

We could take moments around any of the marked points A-E on the diagram. However, we’re going to take moments around B as it enables us to ignore the upward reaction force acting on the rule at B. (This force is not shown on the diagram.)

To indicate that we’re going to be considering the sum of the clockwise moments about point B, we use this intuitive notation:

If we consider the sum of anticlockwise moments about point B, we use this:

We lay out our calculations of the total clockwise and anticlockwise moments about B as follows.

We show that we are going to apply the Principle of Moments (the sum of clockwise moments is equal to the sum of anticlockwise moments for an object in equilibrium) like this:

The rest, as they say, is not history but algebra:

I hope you find this ‘momentary’ convention useful(!)

## Introducing vectors (part 1)

I think that teaching vectors to 14-16 year olds is a bit like teaching them to play the flute; that is to say, it’s a bit like teaching them to play the flute as presented by Monty Python (!)

Monty Python (1972), ‘How to play the flute’

Part of the trouble is that the definition of a vector is so deceptively and seductively easy: a vector is a quantity that has both magnitude and direction.

There — how difficult can the rest of it be? Sadly, there’s a good deal more to vectors than that, just as there’s much more to playing the flute than ‘moving your fingers up and down the outside'(!)

What follows is a suggested outline teaching schema, with some selected resources.

### Resultant vector = total vector: the ‘I’ phase

‘2 + 2 = 4’ is often touted as a statement that is always obviously and self-evidently true. And so it is — arithmetically and for mere scalar quantities. In fact, it would be more precisely rendered as ‘scalar 2 + scalar 2 = scalar 4’.

However, for vector quantities, things are a wee bit different. For vectors, it is better to say that ‘vector 2 + vector 2 = a vector quantity with a magnitude somewhere between 0 and 4’.

For example, if you take two steps north and then a further two steps north then you end up four steps away from where you started. Also, if you take two steps north and then two steps south, then you end up . . . zero steps from where you started.

So much for the ‘zero’ and ‘four’ magnitudes. But where do the ‘inbetween’ values come from?

Simples! Imagine taking two steps north and then two steps east — where would you end up? In other words, what distance and (since we’re talking about vectors) in what direction would you be from your starting point?

This is most easily answered using a scale diagram.

To calculate the vector distance (aka displacement) we draw a line from the Start to the End and measure its length.

The length of the line is 2.8 cm which means that if we walk 2 steps north and 2 steps east then we up a total vector distance of 2.8 steps away from the Start.

But what about direction? Because we are dealing with vector quantities, direction just as important as magnitude. We draw an arrowhead on the purple line to emphasise this.

Students may guess that the direction of the purple ‘resultant’ vector (that is to say, it is the result of adding two vectors) is precisely north-east, but this can be a vague description so let’s use a protractor so that we can find the compass bearing.

And thus we find that the total resultant vector — the result of adding 2 steps north and 2 steps east — is a displacement of 2.8 steps on a compass bearing of 045 degrees.

### Resultant vector = total vector: the ‘We’ phase

How would we go about finding the resultant vector if we moved 3 metres north and 4 metres east? If you have access to an interactive whiteboard, you could choose to use this Jamboard for this phase. (One minor inconvenience: you would have to draw the straight lines freehand but you can use the moveable and rotatable ruler and protractor to make measurements ‘live’ with your class.)

We go through a process similar to the one outlined above.

• What would be a suitable scale?
• How long should the vertical arrow be?
• How long should the horizontal arrow be?
• Where should we place the ‘End’ point?
• How do we draw the ‘resultant’ vector?
• What do we mean by ‘resultant vector’?
• How should we show the direction of the resultant vector?
• How do we find its length?
• How do we convert the length of the arrow on the scale diagram into the magnitude of the displacement in real life?

The resultant vector is, of course, 5.0 m at a compass bearing of 053 degrees.

### Resultant vector = total vector: the ‘You’ phase

Students can complete the questions on the worksheets which can be printed from the PowerPoint below.

Answers are shown on this second PowerPoint, plus an optional digital ruler and protractor are included on the third slide if you wish to use them.

Enjoy!

Uncategorized

## Using dimensional analysis to estimate the energy released by an atomic bomb

Legend has it that in the early 1950s, British physicist G. I. Taylor was visited by some very serious men from the military authorities. His crime? He had apparently secured unauthorised access to worryingly accurate and top secret information about the energy released by the first atom bomb.

Taylor explained that, actually, he hadn’t: he had estimated the energy yield from a series of photographs of the first atomic test explosion published by Life magazine. Taylor had used the standard physics technique known as dimensional analysis.

The published pictures had helpfully included a scale to indicate the size of the atomic fireball in each photograph and Taylor had been able to complete a back-of-the-envelope calculation which gave a surprisingly accurate value for what was then the still highly classified energy yield of an atomic weapon.

This story was shared by the excellent David Cotton (@NewmanPhysics) on Twitter, and included a link to a useful summary which forms the basis of what follows. (NB Any errors or omissions are my own.)

It is presented here for A-level Physics teachers to consider using as an example of the power of dimensional analysis beyond the usual “predicting the form of the equation for the period of a simple pendulum”(!)

### Taylor’s method: step one

Taylor began by assuming that the radius R of the fireball would depend on:

• The energy E released by the bomb. The larger the energy released then the larger the fireball.
• The density of the air ρ. The greater the density of the air then the smaller the fireball since more work would have to be done to push the air out of the path of the fireball.
• The time elapsed t from the explosion. The longer the time then the larger the size of the fireball (until the moment when it began to collapse).

These three factors can be combined into a single relationship:

k is an unknown arbitrary constant. Note that we would expect the exponent y to be negative since R is expected to decrease as ρ increases. We would, however, expect x and z to be positive.

### Taylor’s method: step two

Next we think of the dimensions of each of the values in terms of the basic dimensions or measurements of length [L], mass [M] and time [T].

• R has the dimension of length, so R = [L].
• E is in joules or newton metres (since work done = force x distance). From F=ma we can conclude that the dimensions of newtons are [M] [L] [T]-2. This makes the dimensions of energy [M] [L]2 [T]-2.
• ρ is in kilograms per cubic metre so it has the dimensions [M] [L]-3.
• t has the dimension of time [T].

### Taylor’s method: step three

Next we write equation 1 in terms of the dimensions of each of the quantities. We can ignore k as we assume that this is a purely numerical value with no units. This gives us:

Simplifying this expression, we get:

### Taylor’s method: step four

Next, let’s look at the exponents of [M], [L] and [T].

Firstly, we can see that x + y = 0 since there is no [M] term on the left hand side.

Secondly, we can see that 2x – 3y = 1 since there is an [L] term on the left hand side.

Thirdly, we can see that z – 2x = 0 since there is no [T] term on the left hand side.

### Taylor’s method: step five

We now have a system of three equations detailing three unknowns.

We can solve for x, y and z using simultaneous equations. This gives us x=(1/5), y=(-1/5) and z=(2/5).

### Taylor’s method: step six

Let’s rewrite equation 1 using these values. This gives us:

Rearranging for E gives us:

### Taylor’s method: step seven

Next we read off the value of t=0.006 s and estimate R=75 m from the photograph. The density of air ρ at normal atmospheric pressure is ρ=1.2 kg/m3.

If we substitute these values into equation 6 (assuming that k=1) we get E= 7.9 x 1013 joules.

### Conclusion

Modern sources estimate the yield of the Trinity test as being equivalent to between 18-20 kilotons of TNT. Let’s take the mean value of 19 kilotons. One kiloton is equivalent to 4.184 terajoules. This means that, according to declassified sources that were not available to Taylor, the energy released by the Trinity test was 7.9 x 1013 joules.

As you can see, Taylor’s “guesstimated” value using the dimensional analysis technique was remarkably close to the actual value. No wonder that the military authorities were concerned about this apparent “leak” of classified information.

## Through a glass, lightly

Any sufficiently advanced technology is indistinguishable from magic.

Profiles of the Future (1962)

So wrote Arthur C. Clarke, science fiction author and the man who invented the geosynchronous communications satellite.

Clarke later joked of his regret at the billions of dollars he had lost by not patenting the idea, but one gets the impression that what really rankled him was another (and, to be fair, uncharacteristic) failure of imagination: he did not foresee how small and powerful solid state electronics would become. He had pictured swarms of astronauts crewing vast orbital structures, having their work cut out as they strove to maintain and replace the thousands of thermionic valves burning out under the weight of the radio traffic from Earth . . .

A communications nexus that could fit into a volume the size of a minibus, then (as the technology developed) a suitcase and then a pocket seemed implausible to him — and to pretty much everyone else as well.

And yet, here we are, living in the world that microelectronics has wrought. We are truly in an age of Clarkeian magic: our technology has become so powerful, so reliable and so ubiquitous — and so few of us have a full understanding of how it actually works — that it is very nearly indistinguishable from magic.

I want to outline a fictional vignette on the same theme which Clarke wrote in 1961 which has stayed with me since I read it. Please bear with me while I sketch out the story — it will lend some perspective to what follows.

In the novel A Fall of Moondust (1961) set in the year 2100, the lunar surface transport Selene has become trapped under fifteen metres of moondust. Rescue teams on the surface have managed to drill down to the stricken vessel with a metal pipe to supply the unfortunate passengers and crew with oxygen from the rescue ship. Communication would seem to be impossible as the Selene’s radio has been destroyed, but luckily Chief Engineer (Earthside) Lawrence has a plan:

They would hear his probe, but there was no way in which they could communicate with him. But of course there was. The easiest and most primitive means of all, which could be so readily overlooked after a century and a half of electronics.

A few hours later, the rescue team’s pipe drills through the roof of the sunken vessel.

The brief rush of air gave everyone a moment of unnecessary panic as the pressure equalised. Then the pipe was open to the upper world, and twenty two anxious men and women waited for the first breath of oxygen to come gushing down it.

Instead, the tube spoke.

Out of the open orifice came a voice, hollow and sepulchral, but perfectly clear. It was so loud, and so utterly unexpected, that a gasp of surprise came from the company. Probably not more than half a dozen of these men and women had ever heard of a ‘speaking tube’; they had grown up in the belief that only through electronics could the voice be sent across space. This antique revival was as much a novelty to them as a telephone would have been to an ancient Greek.

### The humble converging lens as an ‘antique revival’

If you hold a magnifying glass (or any converging lens) in front of a white screen, then it will produce a real, inverted image of any bright objects in front of it. This simple act can, believe it or not, draw gasps of surprise from groups of our ‘digital native’ students: they assume that images can only be captured electronically. The fact that a shaped piece of glass can do so is as much a novelty to them as an LCD screen would have been to Galileo.

Think about it for a moment: how often has one of our students seen an image projected by a lens onto a passive screen? The answer is: possibly never.

The cinema? Not necessarily — many cinemas use large electronic screens now; there is no projection room, no projector painting the action on the screen from behind us with ghostly, dancing fingers of light. School? In the past, we had overhead projectors and even interactive whiteboards had lens systems, but these have largely been replaced by LED and LCD screens.

I believe that if you do not take the time to show the phenomenon of a single converging lens projecting a real image on to a passive white screen to your students, they are likely to have no familiar point of reference on which to build their understanding and lens diagrams will remain a puzzling set of lines that have little or no connection to their world.

### Teaching Ray Diagrams

Start with a slide that looks something like this:

What represents the lens? The answer is not the blue oval. On ray diagrams, the lens is represented by the vertical dotted line. F1 and F2 are the focal points of this converging lens and they are each a distance f from the centre of the lens, where f is the focal length.

Now what happens to a light ray from the object that passes through the optical centre of the lens?

The answer, of course, is a big fat nothing. Light rays which pass through the optical centre of a thin lens are undeviated.

Now let’s track what happens to a light ray that travels parallel to the principal axis as shown?

Make sure that your students are aware that this light ray hasn’t ‘missed’ the lens. The lens is the vertical dotted line, not the blue oval. What will happen is that it will be deviated so that it passes through F1 (this is because this is a converging lens; if it had been a diverging lens then it would be be bent so that it appeared to come from F2).

The image is formed where the two light rays cross, as shown below.

We can see that the image is inverted and reduced.

The image is formed close to F1 but not precisely at F1. This is because, although the object is distant from the lens (‘distant’ in this case being ‘further than 2f away’) it is not infinitely far away. However, the further we move the object away from the lens, then the closer to F1 the image is formed. The image will be formed a distance f from the screen when the object in very, very, very large distance away — or an ‘infinite distance’ away, if you prefer.

One of my physics teachers liked to say that ‘Infinity starts at the window sill’. In the context of thinking about lenses, I think he was right . . .

### Some free stuff . . .

The PowerPoint that I used to produce the ray diagrams above is here. It is imperfect in a lot of ways but. truth be told, it has served me well over a number of years. It also features some other slides and animations that you may find useful — enjoy!

### Postscript: ‘Through a glass, darkly’

The phrase ‘Through a glass, darkly’ comes from the writings of the apostle Paul:

For now we see through a glass, darkly; but then face to face: now I know in part; but then shall I know even as also I am known.

KJV 1 Corinthians 13:12

The New International Version translates the phrase less poetically as ‘Now we see but a poor reflection as in a mirror.’

It has been argued that the ‘glass’ Paul was referring to were pieces of naturally-occurring semi-transparent mineral that were used in the ancient world as lenses or windows. They tended to produce a recognisable but distorted view of the world — hence, ‘darkly’.

Better technology means that there is much less distortion produced by our glasses — hence, ‘through a glass, lightly’.